STRENGTH  OF  MATERIALS 


A.  COMPREHENSIVE    PRESENTATION   OF   SCIENTIFIC    METHODS 

OF    LOCATING    AND    DETERMINING    STRESSES    AND 

CALCULATING  THE  REQUIRED  STRENGTH 

AND  DIMENSIONS  OF  BUILDING 

MATERIALS 


EDWARD  R.  MAURER,  B.C.E. 

MECHANICS, 


FBornssoB  OF 


UNIVERSITY    OF    WISCONSIN 


ILLUSTRATED 


AMERICAN  TECHNICAL  SOCIETY 

CHICAGO 

1917 


COPYRIGHT,  1907,  1914,  1917,  BY 
AMERICAN  TECHNICAL  SOCIETY 


COPYRIGHTED    IN    GREAT    BRITAIN 
ALL   RIGHTS    RESERVED 


INTRODUCTION 

EVERY  layman  is  fascinated  by  a  great  engineering  work — a 
large  sewage  system,  a  Keokuk  dam,  a  twenty-story  steel 
structure — and  wishes  he  might  be  able  to  construct  such  work 
and  carry  it  to  completion.  And  yet  he  hardly  appreciates  the 
knowledge,  experience,  and  judgment  necessary  to  bring  such  a 
work  to  a  satisfactory  close.  Time  was  when  all  of  the  details  of 
such  structures  were  determined  by  guesswork,  but  the  develop- 
ments in  science  and  mathematics  have  changed  all  that.  For- 
mulas for  the  various  types  of  stresses  have  been  worked  out; 
constants  for  every  known  material  have  been  collected;  and  a 
multitude  of  diagrams  and  tables  contribute  to  making  the  engi- 
neer's work  as  precise  as  a  bookkeeper's  balance.  The  strength 
and  size  of  every  rivet  and  the  length  and  cross-section  of  every 
girder  in  a  steel  structure  are  figured  so  they  will  bear  the  strains 
put  upon  them.  The  design  of  a  masonry  dam,  the  strength  of 
the  concrete  mixture,  and  the  amount  of  steel  reinforcement  are 
all  mathematically  determined  in  order  to  safely  restrain  the  given 
volume  of  water  behind  it. 

<I  The  final  judgment,  therefore,  as  to  the  size  of  every  part  of 
the  structure — must  depend  upon  the  designer's  knowledge  of 
"Strength  of  Materials."  The  treatment  of  Strength  of  Materials, 
as  found  in  standard  textbooks,  is  so  clothed  in  abstruse  mathe- 
matics that  it  is  impossible  for  the  average  trained  man  to  obtain 
a  working  knowledge  of  the  subject.  As  the  subject  is  one  of  the 
most  important  in  any  of  the  engineering  branches,  the  author 
has  thought  it  wise  to  present  it  in  simple  form,  culling  out  all 
material  that  can  not  be  used  in  practical  design  and  analyzing 
the  subject  from  the  theory  through  to  the  practical  formulas 
without  the  use  of  higher  mathematics.  In  fact,  he  has  used  only 
such  mathematics  as  may  be  easily  understood.  While  the  author 
has  designed  this  work  especially  for  home  study  purposes,  the 
material  is  valuable  to  the  college  trained  man  as  well,  as  it  gives 
in  clear,  concise  form  the  principles  which  are  most  used  in  engi- 
neering and  architectural  work.  It  is  the  hope  of  the  publishers 
that  the  book  will  fill  a  place  among  the  useful  reference  works  in 
the  field  of  engineering. 

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CONTENTS 

PAGE 

Simple  stress 1 

Kinds  of  stress 2 

Tension,  compression,  shear 2 

Unit-stress 3 

Deformation 4 

Elasticity 4 

Hooke's  law,  and  elastic  limit 5 

Ultimate  strength 6 

Stress-deformation  diagram 6 

Working  stress  and  strength 8 

Factor  of  safety 8 

Strength  of  materials  in  tension 11 

Strength  of  materials  in  compression 13 

Strength  of  materials  in  shear 15 

Reaction  of  supports 16 

Moment  of  a  force 16 

Principle  of  moments 17 

Kinds  of  beams 19 

Determination  of  reactions  on  beams 19 

External  shear  and  bending  moment 23 

External  shear 23 

Rule  of  signs 23 

Units  for  shears 24 

Shear  diagrams 27 

Maximum  shear 31 

Bending  moment 32 

Moment  diagrams 36 

Maximum  bending  moment '. . .  .  40 

Table  of  maximum  shears,  moments,  etc 41 

Center  of  gravity  and  moment  of  inertia 41 

Center  of  gravity  of  an  area 41 

Principle  of  moments  applied  to  areas 42 

Center  of  gravity  of  built-up  sections 44 

Moment  of  inertia 46 

Table  of  centers  of  gravity  and  moments  of  inertia 52 

Strength  of  beams 52 

Kinds  of  loads 52 

Transverse,  longitudinal,  inclined  forces 52 

Neutral  surface 54 

Neutral  line 54 

Neutral  axis 54 

Stress  at  cross-section 55 

Fibre  stress. . .  .57 


CONTENTS 

Strength  of  beams  (Continued)  PAGE 

Value  of  the  resisting  moment 58 

First  beam  formula 59 

Applications  of  the  first  beam  formula 59 

Laws  of  strength  of  beams 71 

Modulus  of  rupture 72 

Resisting  shear 73 

Second  beam  formula 73 

Horizontal  shear 75 

Design  of  timber  beams .'76 

Kinds  of  loads  and  beams 79 

Flexure  and  tension 79 

Flexure  and  compression 81 

Combined  flexural  and  direct  stress 83 

Strength  of  columns 85 

End  conditions 85 

Classes  of  columns 86 

Cross  sections  of  columns 86 

Radius  of  gyration 86 

Kinds  of  column  loads 88 

Rankine's  column  formula 88 

Graphical  representation  of  column  formulas 93 

Combination  column  formulas 94 

Straight-line  and  Euler  formulas 94 

Parabola-Euler  formulas 98 

Broken  straight-line  formula 101 

Design  of  columns 102 

Strength  of  shafts 105 

Twisting  moment 105 

Torsional  stress 106 

Resisting  moment 107 

Formula  for  the  strength  of  a  shaft 107 

Formula  for  the  power  which  a  shaft  can  transmit 109 

Stiffness  of  rods,  beams,  and  shafts 110 

Coefficient  of  elasticity Ill 

Temperature  stresses 113 

Deflection  of  beams 114 

Twist  of  shafts 116 

Non-elastic  deformation * 117 

Riveted  joints 118 

Kinds  of  joints 118 

Shearing  strength,  or  shearing  value,  of  a  rivet 118 

Bearing  strength,  or  bearing  value,  of  a  plate 119 

Frictional  strength  of  a  joint .- 119 

Tensile  and  compressive  strength  of  riveted  plates 120 

Computation  of  the  strength  of  a  joint 120 

Efficiency  of  a  joint 122 


STRENGTH  OF  MATERIALS. 


PART  I. 


SIMPLE  STRESS. 

i.  Stress.  When  forces  are  applied  to  a  body  they  tend  in  a 
greater  or  less  degree  to  break  it.  Preventing  or  tending  to  pre- 
vent the  rupture,  there  arise,  generally,  forces  between  every  two 
adjacent  parts  of  the  body.  Thus,  when  a 
load  is  suspended  by  means  of  an  iron  rod, 
the  rod  is  subjected  to  a  downward  pull  at 
its  lower  end  and  to  an  upward  pull  at  its 
upper  end,  and  these  two  forces  tend  to  pull 
it  apart.  At  any  cross-section  of  the  rod 
the  iron  on  either  side  "holds  fast"  to  that  on 
the  other,  and  these  forces  which  the  parts 
of  the  rod  exert  upon  each  other  prevent 
the  tearing  of  the  rod.  For  example,  in  Fig. 
1,  let  a  represent  the  rod  and  its  suspended 
load,  1,000  pounds;  then  the  pull  on  the 
lower  end  equals  1,000  pounds.  If  we  neg- 
lect the  weight  of  the  rod,  the  pull  on  the 

o  s. 

upper  end  is  also  1,000  pounds,  as  shown  in 

Fig.  1  (5) ;    and  the    upper  part   A  exerts 

on  the  lower  part  B  an  upward  pull  Q  equal 

to  1,000  pounds,  while  the  lower  part  exerts 

on  the  upper  a  force  P  also  equal  to  1,000  pounds.     These  two 

forces,  P  and  Q,  prevent  rupture  of  the  rod  at  the  "section"  C;  at 

any  other  section  there  are  two  forces  like  P  and  Q  preventing 

rupture  at  that  section. 

By  stress  at  a  section  of  a  body  is  meant  the  force  which  the 
part  of  the  body  on  either  side  of  the  section  exerts  on  the  other. 
Thus,  the  stress  at  the  section  C  (Fig.  1)  is  P  (or  Q),  and  it  equals 
1,000  pounds. 

a.  Stresses  are  usually  expressed  (in  America)  in  pounds, 
sometimes  in  tons.  Thus  the  stress  P  in  the  preceding  article  is 


.  1. 


;      .STRENGTH  OF  MATERIALS 

v**^-*  *•»  3€f''1*1  *  •»      .      •  *  ?'  • 


1,000  pounds,  or  -J  ton.     Notice  that  this  value  has  nothing  to  do 
with  the  size  of  the  cross-section  on  which  the  stress  acts. 

3.  Kinds  of  Stress,  (a)  "When  the  forces  acting  on  a  body 
(as  a  rope  or  rod)  are  such  that  they  tend  to  tear  it,  the  stress  at 
any  cross  -section  is  called  a  tension  or  a  tensile  stress.  The 
stresses  P  and  Q,  of  Fig.  1,  are  tensile  stresses.  Stretched  ropes, 
loaded  "tie  rods"  of  roofs  and  bridges,  etc.,  are  under  tensile  stress. 
(b.)  "When  the  forces  acting  on  a  body  (as  a  short  post,  brick, 

etc.)  are  such  that  they  tend  to 
crush  it,  the  stress  at  any  sec- 
tion at  right  angles  to  the  di- 
rection of  the  crushing  forces  is 
called  a  pressure  or  a  compres- 
sive  stress.  Fig.  2  («)  repre- 
sents a  loaded  post,  and  Fig.  2 
(&)  the  upper  and  lower  parts. 
The  upper  part  presses  down  on 
B,  and  the  lower  part  presses  up 
on  A,  as  shown.  P  or  Q  is 
the  compressive  stress  in  the 
post  at  section  C.  Loaded  posts, 
or  struts,  piers,  etc.,  are  under 
compressive  stress. 

(c.)     "When  the  forces  acting 
on  a  body  (as  a  rivet  in  a  bridge 


T  P 

cu 


Fig.  2. 


joint)  are  such  that  they  tend  to  cut  or  "  shear  "  it  across,  the  stress 
at  a  section  along  which  there  is  a  tendency  to  cut  is  called  a  shear 
or  a  shearing  stress.  This  kind  of  stress  takes  its  name  from  the 
act  of  cutting  with  a  pair  of  shears.  In  a  material  which  is  being 
cut  in  this  way,  the  stresses  that  are  being  "  overcome  "  are  shear- 
ing stresses.  Fig.  3  (&)  represents  a  riveted  joint,  and  Fig.  3  (J>) 
two  parts  of  the  rivet.  The  forces  applied  to  the  joint  are  such 
that  A  tends  to  slide  to  the  left,  and  B  to  the  right;  then  B  exerts 
on  A  a  force  P  toward  the  right,  and  A  on  B  a  force  Q  toward  the 
left  as  shown.  P  or  Q  is  the  shearing  stress  in  the  rivet. 

Tensions,  Compressions  and  Shears  are  called  simple  stresses. 
Forces  may  act  upon  a  body  so  as  to  produce  a  combination  of  simple 
stresses  on  some  section  ;  such  a  combination  is  called  a  complex 


STRENGTH  OF  MATERIALS  3 

stress.   The  stresses  in  beams  are  usually  complex.   There  are  other 
terms  used  to  describe  stress;  they  will  be  defined  farther  on. 

4.  Unit=Stress.  It  is  often  necessary  to  specify  not  merely 
the  amount  of  the  entire  stress  which  acts  on  an  area,  but  also  the 
amount  which  acts  on  each  unit  of  area  (square  inch  for  example). 
By  unit-stress  is  meant  stress  per  unit  area. 

To  find  the  value  of  a  unit-stress:  Divide  the  whole  stress  by 
the  whole  area  of  the  section  on  which  it  acts,  or  over  which  it  is 
distributed.  Thus,  let 

P  denote  the  value  of  the  whole  stress, 

A  the  area  on  which  it  acts,  and 

S  the  value  of  the  unit-stress;  then 


=  AS.  (l) 

Strictly  these  formulas  apply  only  when  the  stress  P  is  uniform, 


Fig.  3. 

that  is,  when  it  is  uniformly  distributed  over  the  area,  each  square 
inch  for  example  sustaining  the  same  amount  of  stress.  When 
the  stress  is  not  uniform,  that  is,  when  the  stresses  on  different 
square  inches  are  not  equal,  then  P-r-A  equals  the  average  value 
of  the  unit-stress. 

5.  Unit-stresses  are  usually  expressed  (in  America)  in 
pounds  per  square  inch,  sometimes  in  tons  per  square  inch.  If 
P  and  A  in  equation  1  are  expressed  in  pounds  and  square 
inches  respectively,  then  S  will  be  in  pounds  per  square  inch;  and 
if  P  and  A  are  expressed  in  tons  and  square  inches,  S  will  be  in 
tons  per  square  inch. 

Examples.  1.  Suppose  that  the  rod  sustaining  the  load  in 
Fig.  1  is  2  square  inches  in  cross-section,  and  that  the  load  weighs 
1,000  pounds.  What  is  the  value  of  the  unit-strese  ? 


4  STRENGTH  OF  MATERIALS 

Here  P  =  1,000  pounds,  A=  2  square  inches;  hence. 
S  =  -— — •  =  500  pounds  per  square  inch. 

2.  Suppose  that  the  rod  is  one-half  square  inch  in  cross -sec- 
tion. What  is  the  value  of  the  unit-stress  ? 

A  =  -^-square  inch,  and,  as  before,  P==  1,000  pounds;  hence 

S  =  1,000-J — g-  =  2,000  pounds  per  square  inch. 

Notice  that  one  must  always  divide  the  whole  stress  by  the  area  to  get 
the  unit-stress,  whether  the  area  is  greater  or  less  than  one. 

6.  Deformation.     Whenever  forces  are  applied  to  a  body  it 
changes  in  size,  and  usually  in  shape  also.     This  change  of  size 
and  shape  is  called  deformation.     Deformations  are  usually  meas- 
ured in  inches;  thus,  if  a  rod  is  stretched  2  inches,  the  "elonga- 
tion"=  2  inches. 

7.  Unit-Deformation.     It  is  sometimes  necessary  to  specify 
not  merely  the  value  of  a  total  deformation  but  its  amount  per 
unit  length  of  the  deformed  body.     Deformation  per  unit  length 
of  the  deformed  body  is  called  unit-deformation. 

To  find  the  value  of  a  unit-deformation :     Divide  the  whole 
deformation  1y  the  length  over  which  it  is  distributed.     Thus,  if 
D  denotes  the  value  of  a  deformation, 
I    the  length, 
s    the  unit-deformation,  then 

8=-j-,  also  T>=ls.  (2) 

Both  D  and  I  should  always  be  expressed  in  the  same  unit. 

Example.  Suppose  that  a  4-foot  rod  is  elongated  \  inch. 
What  is  the  value  of  the  unit-deformation? 

Here  D— \  inch,  and  1=4:  feet=48  inches; 
hence  «=J-j-48=-J^r  inch  per  inch. 
That  is,  each  inch  \s  elongated  -£v  inch. 

Unit-elongations  are  sometimes  expressed  in  per  cent.  To 
express  a**  elongation  in  per  cent:  Divide  the  elongation  in  inches 
by  the  original  length  in  inches,  and  multiply  l)y  100. 

8.  Elasticity.     Most  solid  bodies  when  deformed  will  regain 
more  or  less  completely  their  natural  size  and  shape  when  the  de. 


STRENGTH  OF  MATERIALS  5 

forming  forces  cease  to  act.     This  property  of  regaining  size  and 
shape  is  called  elasticity. 

"We  may  classify  bodies  into  kinds  depending  on  the  degree 
of  elasticity  which  they  have,  thus : 

1.  Perfectly  elastic   bodies;    these  will  regain   their  orig- 
inal form  and  size  no  matter  how  large  the  applied  forces  are  if 
less  than  breaking  values.     Strictly  there  are  no  such  materials, 
but  rubber,  practically,  is  perfectly  elastic. 

2.  Imperfectly  elastic  bodies;  these  will  fully  regain  their 
original  form  and  size  if  the  applied  forces  are  not  too  large,  and 
practically  even  if  the  loads  are  large  but  less  than  the  breaking 
value.     Most  of  the  constructive  materials  belong  to  this  class. 

3.  Inelastic  or  plastic  bodies;  these  will  not  regain  in  the 
least  their  original  form  when  the  applied  forces  cease  to  act.     Clay 
and  putty  are  good  examples  of  this  class. 

9.  Hooke's  Law,  and  Elastic  Limit.  If  a  gradually  increas- 
ing force  is  applied  to  a  perfectly  elastic  material,  the  deformation 
increases  proportionally  to  the  force;  that  is,  if  P  and  P'  denote 
two  values  of  the  force  (or  stress),  and  D  and  D'  the  values  of  the 
deformation  produced  by  the  force, 

thenP:P'::D:D'. 

This  relation  is  also  true  for  imperfectly  elastic  materials, 
provided  that  the  loads  P  and  P'  do  not  exceed  a  certain  limit  depend- 
ing on  the  material.  Beyond  this  limit,  the  deformation  increases 
much  faster  than  the  load;  that  is,  if  within  the  limit  an  addition 
of  1,000  pounds  to  the  load  produces  a  stretch  of  0.01  inch,  beyond 
the  limit  an  equal  addition  produces  a  stretch  larger  and  usually 
much  larger  than  0.01  inch. 

Beyond  this  limit  of  proportionality  a  part  of  the  deformation 
is  permanent;  that  is,  if  the  load  is  removed  the  body  only  partially 
recovers  its  form  and  size.  The  permanent  part  of  a  deformation 
is  called  set. 

The  fact  that  for  most  materials  the  deformation  is  propor- 
tional to  the  load  within  certain  limits,  is  known  as  Hooke's  Law. 
The  unit-stress  within  which  Hooke's  law  holds,  or  above  which 
the  deformation  is  not  proportional  to  the  load  or  stress,  is  called 
elastic  limit. 


6  STRENGTH  OF  MATERIALS 

10.  Ultimate  Strength.     By  ultimate  tensile,  compressive, 
or  shearing  strength  of  a  material  is  meant    the  greatest  tensile, 
compressive,  or  shearing  unit-stress  which  it  can  withstand. 

As  before  mentioned,  when  a  material  is  subjected  to  an  in- 
creasing load  the  deformation  increases  faster  than  the  load  beyond 
the  elastic  limit,  and  much  faster  near  the  stage  of  rupture.  Not 
only  do  tension  bars  and  compression  blocks  elongate  and  shorten 
respectively,  but  their  cross -sectional  areas  change  also;  tension 
bars  thin  down  and  compression  blocks  "swell  out"  more  or  less. 
The  value  of  the  ultimate  strength  for  any  material  is  ascertained 
by  subjecting  a  specimen  to  a  gradually  increasing  tensile,  com- 
pressive,  or  shearing  stress,  as  the  case  may  be,  until  rupture  oc- 
curs, and  measuring  the  greatest  load.  The  breaking  load  divided 
l)y  the  area  of  the  original  cross -section  sustaining  the  stress,  is  the 
value  of  the  ultimate  strength. 

Example.  Suppose  that  in  a  tension  test  of  a  wrought-iron 
rod  -|  inch  in  diameter  the  greatest  load  was  12,540  pounds.  "What 
is  the  value  of  the  ultimate  strength  of  that  grade  of  wrought  iron? 

The  original  area  of  the  cross-section  of  the  rod  was 
0.7854  (diameter)2=0.7854x  J=0.1964  square  inches;  hence 
the  ultimate  strength  equals 

12,540-^0.1964:=63,850  pounds  per  square  inch,  (nearly). 

11.  Stress- Deformation    Diagram.     A  "test"  to  determine 
the  elastic  limit,  ultimate  strength,  and  other  information  in  re- 
gard to  a  material  is  conducted  by  applying  a  gradually  increasing 
load  until  the  specimen  is  broken,  and  noting  the  deformation  cor- 
responding to  many  values  of  the  load.     The  first  and  second  col- 
umns of  the  following  table  are  a  record  of  a  tension  test  on  a  steel 
rod  one  inch  in  diameter.     The  numbers  in  the  first  column  are 
the  values  of  the  pull,  or  the  loads,  at  which  the  elongation  of 
the  specimen  was  measured.     The  elongations  are  given  in  the  sec- 
ond column.     The  numbers  in  the  third  and  fourth  columns  are 
the  values  of  the  unit-stress  and  unit-elongation  corresponding  to 
the    values  of  the  load  opposite  to  them.     The  numbers  in  the 
third  column  were  obtained  from  those  in  the  first  by  dividing 
the    latter  by  the   area  of   the  cross-section    of   the   rod,  0.7854 
square  inches.     Thus, 

3,930-^0.7854=5,000 
7,850-^0.7854=10,000,  etc. 


STRENGTH  OF  MATERIALS 


Total  Pull 
iu  pounds,  P 

Deformation 
in  inches,  D 

Unit-Stress  in 
pounds  per 
square  inch,  S 

Unit- 
Deformation, 
s 

3930 

0.00136 

5000 

0.00017 

7850 

.00280 

10000 

.00035 

11780 

.00404 

15000 

.00050 

15710 

.00538 

20000 

.00067 

19635 

.00672 

25000 

.00084 

23560 

.00805 

30000 

.00101 

27490 

.00942 

35000 

.00118 

31415 

.01080 

40000 

.00135 

35345 

.01221 

45000 

.00153 

39270 

.0144 

50000 

.00180 

43200 

.0800 

55000 

.0100 

47125 

.1622 

60000 

.0202 

51050 

.201 

65000 

.0251 

54980 

.281 

70000 

.0351 

58910 

.384 

75000 

.048 

62832 

.560 

80000 

.070 

65200 

1.600 

83000 

.200 

The  numbers  in  the  fourth  column  were  obtained  by  dividing 
those  in  the  second  by  the  length  of  the  specimen  (or  rather  the 
length  of  that  part  whose  elongation  was  measured),  8  inches. 
Thus, 

0.00136 -^-8  =  0.00017, 
.00280-^8  =    .00035,  etc. 

Looking  at  the  first .  two  columns  it  will  be  seen  that  the  elonga- 
tions are  practically  proportional  to  the  loads  up  to  the  ninth  load, 
the  increase  of  stretch  for  each  increase  in  load  being  about  0.00135 
inch;  but  beyond  the  ninth  load  the  increases  of  stretch  are  much 
greater.  Hence  the  elastic  limit  was  reached  at  about  the  ninth 
load,  and  its  value  is  about  45,000  pounds  per  square  inch.  The 
greatest  load  was  65,200  pounds,  and  the  corresponding  unit-stress, 
83,000  pounds  per  square  inch,  is  the  ultimate  strength. 

Nearly  all  the  information  revealed  by  such  a  test  can  be 
well  represented  in  a  diagram  called  a  stress-deformation  diagram. 
It  is  made  as  follows:  Lay  off  the  values  of  the  unit-deformation 
(fourth  column)  along  a  horizontal  line,  according  to  some  con- 
venient scale,  from  some  fixed  point  in  the  line.  At  the  points  on 
the  horizontal  line  representing  the  various  unit-elongations,  lay 
off  perpendicular  distances  equal  to  the  corresponding  unit-stresses. 
Then  connect  by  a  smooth  curve  the  upper  ends  of  all  those  dis- 
tances, last  distances  laid  off.  Thus,  for  instance,  the  highest  unit- 


8  STRENGTH  OF  MATERIALS 

elongation  (0.20)  laid  off  from  o  (Fig.  4)  fixes  the  point  a,  and  a 
perpendicular  distance  to  represent  the  highest  unit-  stress  (83,000) 
fixes  the  point  1).  All  the  points  so  laid  off  give  the  curve  ocb.  The 
part  ocy  within  the  elastic  limit,  is  straight  and  nearly  vertical 
while  the  remainder  is  curved  and  more  or  less  horizontal,  especially 
toward  the  point  of  rupture  t>.  Fig.  5  is  a  typical  stress-defor- 
mation diagram  for  timber,  cast  iron,  wrought  iron,  soft  and  hard 
steel,  in  tension  and  compression. 

12.  Working  Stress  and  Strength,  and  Factor  of  Safety. 
The  greatest  unit-stress  in  any  part  of  a  structure  when  it  is  sus- 

taining its  loads  is  called  the 
working  stress  of  that  part.  If 
it  is  under  tension,  compression 
and  shearing  stresses,  then  the 
corresponding  highest  unit- 
stresses  in  it  are  called  its  work- 
ing stress  in  tension,  in  com- 
pression, and  in  shear  respect- 
ively; that  is,  we  speak  of  as 
~  """  many  working  stresses  as  it  has 

kinds  of  stress. 

By  working  strength  of  a  material  to  be  used  for  a  certain 
purpose  is  meant  the  highest  unit-  stress  to  which  the  material 
ought  to  be  subjected  when  so  used.  Each  material  has  a  working 
strength  for  tension,  for  compression,  and  for  shear,  and  they  are  in 
general  different. 

By  factor  of  safety  is  meant  the  ratio  of  the  ultimate  strength 
of  a  material  to  its  working  stress  or  strength.  Thus,  if 

O  O  ' 

Su  denotes  ultimate  strength, 
Sw  denotes  working  stress  or  strength,  and 
f  denotes  factor  of  safety,  then 


When  a  structure  which  has  to  stand  certain  loads  is  about 
to  be  designed,  it  is  necessary  to  select  working  strengths  or  fac- 
tors of  safety  for  the  materials  to  be  used.  Often  the  selection  is 
a  matter  of  great  importance,  and  can  be  wisely  performed  only 
by  an  experienced  engineer,  for  this  is  a  matter  where  hard  -and- 


STRENGTH  OF  MATERIALS  9 

fast  rules  should  not  govern  but  rather  the  judgment  of  the  expert. 
But  there  are  certain  principles  to  be  used  as  guides  in  making  a 
selection,  chief  among  which  are: 

1.  The  working  strength  should  be  considerably  below  the 
elastic  limit.  (Then  the  deformations  will  bft  small  and  not  per- 
manent.) 


Fig.  5.     (After  Johnson.) 

2.  The  working  strength  should  be  smaller  for  parts  of  a 
structure  sustaining  varying  loads  than  for  those  whose  loads  are 
steady.       (Actual  experiments  have  disclosed   the  fact  that  the 
strength  of  a  specimen  depends  on  the  kind  of  load  put  upon  it, 
and  that  in  a  general  way  it  is  less  the  less  steady  the  load  is.) 

3.  The  working  strength  must  be  taken  low  for  non -uniform 
material,   where  poor  workmanship  may  be  expected,  when  the 


10 


STRENGTH  OF  MATERIALS 


loads  are  uncertain,  etc.  Principles  1  and  2  have  been  reduced 
to  figures  or  formulas  for  many  particular  cases,  but  the  third  must 
remain  a  subject  for  display  of  judgment,  and  even  good  guessing 
in  many  cases. 

The  following  is  a  table  of  factors  of  safety*  which  will  be 
used  in  the  problems: 

Factors  of  Safety. 


Materials. 

For  steady 
stress. 
(Buildings.) 

For  varying 
stress. 
(Bridges.) 

For  shocks. 
(Machines.) 

Timber 
Brick  and  stone 
Cast  iron 
Wrought  iron 
Steel 

8 
15 
6 
4 
5 

10 
25 
15 
6 

7 

15 
30 
20 
10 
15 

They  must  be  regarded  as  average  values  and  are  not  to  be 
adopted  in  every  case  in  practice. 

Examples.  1.  A  wrought -iron  rod  1  inch  in  diameter  sus- 
tains a  load  of  30,000  pounds.  "What  is  its  working  stress?  If 
its  ultimate  strength  is  50rOOO  pounds  per  square  inch,  what  is 
its  factor  of  safety  ? 

The  area  of  the  cross- section  of  the  rod  equals  0.7854  X  (diam- 
eter ) 2— 0.7854  X  12=0.7854  square  inches.  Since  the  whole  stress 
on  the  cross-section  is  30,000  pounds,  equation  1  gives  for  the 
unit  working  stress 

30  000 
S  =  ~  VGKA  —  38,197  pounds  per  square  inch. 

Equation  3  gives  for  factor  of  safety 

50,000 


/-a 


__  -i    o 


38,197 

2.  How  large  a  steel  bar  or  rod  is  needed  to  sustain  a  steady 
pull  of  100,000  pounds  if  the  ultimate  strength  of  the  material  is 
65,000  pounds  ? 

The  load  being  steady,  we  use  a  factor  of  safety  of  5  (see  table 
above);  hence  the  working  strength  to  be  used  (see  equation  3)  is 

S  = — ^—     =  13,000  pounds  per  square  inch. 

The  proper  area  of  the  cross -section  of  the  rod  can   now   be  com- 
puted from  equation  1  thus: 
*Taken  from  Merriman's  "Mechanics  of  Materials/' 


STRENGTH  OF  MATERIALS  11 


P      100,000 
A  =  g-  -  =  -13  '        =  7.692  square  inches. 


A  bar  2x4  inches  in  cross-section  would  be  a  little  stronger 
than  necessary.  To  find  the  diameter  (d)  of  a  round  rod  of  suffi- 
cient strength,  we  write  0.7854  d2  =  7.692,  and  solve  the  equation 
fore?/  thus: 

7  fiQ2 
d2=^j^=  9.794,  or  rf  =  3,129  inches. 

3.  How  large  a  steady  load  can  a  short  timber  post  safely  sus- 
tain if  it  is  10x10  inches  in  cross-section  and  its  ultimate  com- 
pressive  strength  is  10,000  pounds  per  square  inch  ? 

According  to  the  table  (page  12)  the  proper  factor  of  safety  is 
8,  and  hence  the  working  strength  according  to  equation  3  is 

10,000 
S  =  —  -K  —  =  1,250  pounds  per  square  inch. 

The  area  of  the  cross-section  is  100  square  inches;  hence  the  safe 
load  (see  equation  1)  is 

P  =  100  X  1,250  =  125,000  pounds. 

4.  When  a  hole  is  punched   through  a  plate  the  shearing 
strength  of  the  material  has  to  be  overcome.     If  the  ultimate  shear- 

o 

ing  strength  is  50,000  pounds  per  square  inch,  the  thickness  of  the 
plate  -J  inch,  and  the  diameter  of  the  hole  f  inch,  what  is  the  value 
of  the  force  to  be  overcome  ? 

The  area  shorn  is  that  of  the  cylindrical  surface  of  the  hole 
or  the  metal  punched  out^  that  is 

3.1416  X  diameter  X  thickness  =  3.1416  X  f  X  J  =  1.178  sq.  in. 
Hence,  by  equation  1,   the  total  shearing  strength  or  resistance 
to  punching  is 

P  =  1.178  X  50,000  =  58,900  pounds. 

STRENGTH  OF  MATERIALS  UNDER  SIMPLE  STRESS. 

13.  Materials   in   Tension.     Practically  the  only  materials 
used  extensively  under  tension  are  timber,  wrought  iron  and  steel, 
and  to  some  extent  cast  iron. 

14.  Timber.    A  successful  tension  test  of  wood  is  difficult, 
as  the  specimen  usually  crushes  at  the  ends  when  held  in  the  test- 
ing machine,  splits,  or  fails  otherwise  than  as  desired.     Hence  the 


12  STRENGTH  OF  MATERIALS 

tensile  strengths  of  woods  are  not  well  known,  but  the  following 
may  be  taken  as  approximate  average  values  of  the  ultimate 
strengths  of  the  woods  named,  when  "dry  out  of  doors." 

Hemlock,  7,000  pounds  per  square  inch. 

White  pine,  8,000  "  "    . 

Yellow  pine,  long  leaf,  12,000  " 

"          "   ,  short  leaf,  10,000  " 

Douglas  spruce,  10,000  " 

White  oak,  12,000 

Red  oak,  9,000 

15.  Wrought  Iron.     The    process    of   the   manufacture   of 
wrought  iron  gives  it  a  "grain,"  and  its  tensile  strengths  along  and 
across  the  grain  are  unequal,  the  latter  being  about  three-fourths 
of  the  former.     The  ultimate  tensile  strength  of  wrought   iron 
along  the  grain  varies  from  45,000  to  55,000  pounds  per  square 
inch.     Strength  along   the  grain   is  meant  when   not  otherwise 
Btated. 

The  strength  depends  on  the  size  of  the  piece,  it  being  greater 
for  small  than  for  large  rods  or  bars,  and  also  for  thin  than  for 
thick  plates.  The  elastic  limit  varies  from  25,000  to  40,000 
pounds  per  square  inch,  depending  on  the  size  of  the  bar  or  plate 
even  more  than  the  ultimate  strength.  Wrought  iron  is  very 
ductile,  a  specimen  tested  in  tension  to  destruction  elongating  from 
5  to  25  per  cent  of  its  length. 

16.  Steel.     Steel  has  more  or  less  of  a  grain  but  is  practically 
^f  the  same  strength  in  all  directions.     To  suit  different  purposes, 
steel  is  made  of  various  grades,  chief  among  which  may  be  men- 
tioned   rivet  steel,    sheet   steel    (for  boilers),   medium  steel  (for 
bridges  and  buildings),  rail  steel,  tool  and  spring  steel.     In  general, 
these  grades  of  steel  are  hard  and  strong  in  the  order  named,  the 
ultimate  tensile   strength  ranging  from  about   50,000  to  160,000 
pounds  per  square  inch. 

There  are  several  grades   of  structural  steel,  which  may  be 
described  as  follows:* 
1.     Rivet  steel: 

Ultimate  tensile  strength,  48,000  to  58,000  pounds  per  square  inch. 
Elastic  limit,  not  less  than  one-half  the  ultimate  strength. 
Elongation,  26  per  cent. 
Bends  180  degrees  flat  on  itself  without  fracture. 

*Taken  from  "  Manufacturer's  Standard  Specifications." 


STRENGTH  OF  MATERIALS  13 

2.  Soft  steel: 

Ultimate  tensile  strength,  52,000  to  62,000  pounds  per  square  inch. 
Elastic  limit,  not  less  than  one-half  the  ultimate  strength. 
Elongation,  25  per  cent. 
Bends  180  degrees  flat  on  itself. 

3.  Medium  steel: 

Ultimate  tensile  strength,  60,000  to  70,000  pounds  per  square  inch. 
Elastic  limit,  not  less  than  one-half  the  ultimate  strength.  • 
Elongation,  22  per  cent. 

Bends  180  degrees  to  a  diameter  equal  to  the  thickness  of  the 
specimen  without  fracture. 

17.  Cast  Iron.     As   in   the  case    of   steel,   there  are   many 
grades  of  cast  iron.     The  grades  are  riot  the  same  for  all  localities 
or  districts,  but  they  are  based  on  the  appearance  of  the  fractures, 
which  vary  from  coarse  dark  grey  to  fine  silvery  white. 

The  ultimate  tensile  strength  does  not  vary  uniformly  with 
the  grades  but  depends  for  the  most  part  on  the  percentage  of 
"combined  carbon"  present  in  the  iron.  This  strength  varies  from 
15,000  to  35,000  pounds  per  square  inch,  20,000  being  a  fair 
average. 

Cast  iron  has  no  well-defined  elastic  limit  (see  curve  for  cast 
iron,  Fig.  5).  Its  ultimate  elongation  is  about  one  per  cent. 

EXAMPLES  FOR  PRACTICE. 

1.  A  steel  wire  is  one-eighth  inch  in  diameter,  and  the  ulti- 
mate tensile  strength  of  the  material  is  150,000  pounds  per  square 
inch.     How  large  is  its  breaking  load  ?  Ans.     1,845  pounds. 

2.  A  wrought-iron  rod    (ultimate  tensile  strength   50,000 
pounds  per  square  inch)  is  2  inches  in  diameter.       How  large  a 
steady  pull  can  it  safely  bear  ?  Ans.     39,270  pounds. 

18.  Materials  in  Compression.       Unlike    the     tensile,     the 
compressive  strength  of  a  specimen  or  structural  part  depends  on 
its  dimension  in  the    direction  in  which  the  load  is  applied,  for, 
in  compression,  a  long  bar  or  rod  is  weaker  than  a  short  one.     At 
present  we  refer  only  to  the  strength  of  short  pieces  such  as  do 
not  bend  under  the  load,   the  longer  ones    (columns)  being  dis- 
cussed farther  on. 

Different  materials  break  or  fail  under  compression,  in  two 
very  different  ways: 

1.     Ductile  materials  (structural  steel,   wrought   iron,  etc.), 


14  STRENGTH  OF  MATERIALS 

and  wood  compressed  across  the  grain,  do  not  fail  by  breaking  Into 
two  distinct  parts  as  in  tension,  but  the  former  bulge  out  and 
flatten  under  great  loads,  while  wood  splits  and  mashes  down. 
There  is  no  particular  point  or  instant  of  failure  under  increasing 
loads,  and  such  materials  have  no  definite  ultimate  strength  in 
compression. 

2.  Brittle  materials  (brick,  stone,  hard  steel,  cast  iron,  etc.), 
and  wood  compressed  along  the  grain,  do  not  mash  gradually,  but 
fail  suddenly  and  have  a  definite  ultimate  strength  in  compression. 
Although  the  surfaces  of  fracture  are  always  much  inclined  to  the 
direction  in  which  the  load  is  applied  (about  45  degrees),  the  ulti. 
mate  strength  is  computed  by  dividing  the  total  breaking  load  by 
the  cross -sectional  area  of  the  specimen. 

The  principal  materials  used  under  compression  in  structural 
work  are  timber,  wrought  iron,  steel,  cast  iron,  brick  and  stone. 

O 

19.  Timber.     As  before  noted,  timber  has  no  definite  ulti- 
mate compressive  strength  across  the  grain.      The  U.  S.  Forestry 
Division  has  adopted  certain  amounts  of  compressive  deformation 
as  marking  stages  of   failure.      Three   per  cent  compression  is 
regarded  as  "a  working  limit  allowable,"  and  fifteen  per  cent  as 
"an  extreme  limit,  or  as  failure."   The  following  (except  the  first) 
are  values    for  compressive  strength  from  the  Forestry  Division 
Reports,  all  in  pounds  per  square  inch: 

Ultimate  strength         3£  Compression 
along  the  grain.  across  the  grain 

Hemlock 6,000 

White  pine 5,400  700 

Long-leaf  yellow  pine 8,000  1,260 

Short-leaf  yellow  pine 6,500  1,050 

Douglas  spruce 5,700  800 

White  oak 8,500  2,200 

Red  oak 7,200  2,300       . 

20.  Wrought  Iron.     The  elastic  limit  of  wrought  iron,  as  be- 
fore noted,  depends  very  much  upon  the  size  of  the  bars  or  plate,  it 
being  greater  for  small  bars  and  thin  plates.       Its  value  for  com- 
pression is  practically  the  same  as  for  tension,  25,000  to  40,000 
pounds  per  square  inch. 

21.  Steel.       The  hard  steels  have  the    highest    compressive 
strength;  there  is  a  recorded  value  of  nearly  400,000  pounds  per 
square  inch,  but  150,000  is  probably  a  fair  average. 


STRENGTH  OF  MATERIALS  15 

The  elastic  limit  in  compression  is  practically  the  same  as  in 
tension,  which  is  about  60  per  cent  of  the  ultimate  tensile  strength, 
or,  for  structural  steel,  about  25,000  to  42,000  pounds  per  square 

inch. 

22.  Cast  Iron.       This  is  a  very  strong  material  in  compres- 
sion, in  which  way,  principally,  it  is  used  structurally.      Its  ulti- 
mate strength  depends  much  on  the  proportion  of  "combined  car- 
bon" and  silicon  present,  and  varies  from  50,000  to  200,000  pounds 
per  square  inch,   90,000  being  a  fair  average.      As  in  tension, 
there  is  no  well-defined  elastic  limit  in  compression  (see  curve  for 
cast  iron,  Fig.  5). 

23.  Brick.      The  ultimate  strengths  are  as  various  as  the 
kinds  and  makes  of  brick.      For  soft  brick,  the  ultimate  strength 
is  as  low  as  500  pounds  per  square  inch,  and  for  pressed  brick  it 
varies   from   4,000  to  20,000  pounds  per  square  inch,  8,000  to 
10,000  being  a  fair  average.      The  ultimate  strength  of  good  pav- 
ing brick  is  still  higher,  its  average  value  being  from  12,000  to 
15,000  pounds  per  square  inch. 

24.  Stone.       Sandstone,    limestone    and     granite   are     the 
principal  building  stones.      Their  ultimate   strengths  in  pounds 
per  square  inch  are  about  as  follows: 

Sandstone,*  5,000  to  16,000,  average   8,000. 
Limestone,*  8,000  "  16,000,        "       10,000. 
.  Granite,      14,000  "  24,000,        "       16,000. 
Compression  at  right  angles  to  the  "bed"  of  the  stone. 

EXAMPLES  FOR  PRACTICE. 

1.  A  limestone  12  X 12  inches  on  its  bed  is  used  as  a  pier 
cap,  and  bears  a  load  of  120,000  pounds.       What  is  its  factor  of 
safety  ?  Ans.     12. 

2.  How  large  a  post  (short)  is  needed  to  sustain  a  steady 
load  of  100,000  pounds  if  the  ultimate  compressive  strength  of 
the  wood  is  10,000  pounds  per  square  inch  ?       Ans.  8  X 10  inches. 

25.  Materials  in  Shear.     The  principal  materials  used  under 
shearing   stress  are   timber,    wrought  iron,   steel  and  cast  iron. 
Partly    on    account    of   the   difficulty    of   determining   shearing 
strengths,  these  are  not  well  known. 

26.  Timber.     The  ultimate  shearing  strengths  of  the  more 
important  woods  along  the  grain  are  about  as  follows: 


16  STRENGTH  OF  MATERIALS 

• 

Hemlock,  300  pounds  per  square  inch. 

White  pine,  400  "  " 

Long-leaf  yellow  pine,     850  "  " 

Short-leaf       "          "        775  "  " 

Douglas  spruce,  500  "  " 

White  oak,  1,000         •     "  " 

Red  oak,  1,100  "  " 

Wood  rarely  fails  by  shearing  across  the  grain.     Its  ultimate 


Fig.  6  a.  Fig.  6  b. 

shearing  strength  in  that  direction  is  probably  four  or  five  timsa 
the  values  above  given. 

27.  Metals.     The  ultimate  shearing  strength    of   wrought 
iron,  steel,  and  cast  iron  is  about  80  per  cent  of  their  respective 
ultimate  tensile  strengths. 

EXAMPLES  FOR  PRACTICE. 

1.  How  large  a  pressure  P  (Fig.  6  a)  exerted  on  the  shaded 
area  can  the  timber  stand  before  it  will  shear  off  on  the  surface 
abed)  if  ab  =  6  inches  and  "bo  =  10  inches,  and  the  ultimate  shear- 
ing  strength  of  the  timber  is  400  pounds  per  square  inch  ? 

Ans.     24,000  pounds. 

2.  When  a  bolt  is  under  tension,  there  is  a  tendency  to  tear 
the  bolt  and  to  "strip"  or  shear  off  the  head.     The  shorn  area 
would   be   the   surface  of  the  cylindrical  hole  left  in  the  head. 
Compute  the  tensile  and  shearing  unit-stresses  when  P  (Fig.  6  £) 
equals  30,000  pounds,  d  —  2  inches,  and  t  —  3  inches. 

j  Tensile  unit-stress,  9,550  pounds  per  square  inch. 
"  (  Shearing  unit-stress,  1,591  pounds  per  square  inch. 

REACTIONS  OF  SUPPORTS. 

28.  Moment  of  a  Force.     By  moment  of  a  force  with  re- 
epect  to  a  point  is  meant  its  tendency  to  produce  rotation  about 
that  point.     Evidently  the  tendency  depends  on  the  magnitude  of 
the  force  and  on  the  perpendicular  distance  of  the  line  of  action 
of  the  force  from  the  point  :  the  greater  the  force  and  the  per- 
pendicular distance,  the  greater  the  tendency;  hence  the  moment 


STRENGTH  OF  MATERIALS 


17 


of  a  force  with  respect  to  a  point  equals  the  product  of  the  force 
and  the  perpendicular  distance  from  the  force  to  the  point. 

The  point  with  respect  to  which  the  moment  of  one  or  more 
forces  is  taken  is  called  an  origin  or  center  of  moments,  and  the 
perpendicular  distance  from  an  origin  of  moments  to  the  line  of 
action  of  a  force  is  called  the  arm  of  the  force  with  respect  to 
that  origin.  Thus,  if  F1  and  F2  (Fig.  7)  are  forces,  their  arms 
with  respect  to  O'  are  «/  and  a2'  respectively,  and  their  moments 
are  F^'i  and  F2<r'2.  "With  respect  to  O"  their  arms  are  #/'  and  a" 
respectively,  and  their  moments  are  F^/'  and  F2«2". 

If  the  force  is  expressed  in  pounds  and  its  arm  in  feet,  the 
moment  is  in  foot-pounds;  if  the  force  is  in  pounds  and  the  arm 
in  inches,  the  moment  is  in  inch-pounds. 

29.  A  sign  is  given   to  the  moment  of  a  force  for  conven- 
ience;  the  rule   used  herein   is   as  follows:      The  moment  of  a 
force  about  a  point  is  positive  or  negative  according  as  it  tends 
to  turn  the  hody  about  that  point  in  the  clockwise  or  counter- 
clockwise direction*. 

Thus  the  moment  (Fig.  7) 

of  Fj  about  O'  is  negative,  about  O"  positive; 
"  F2      "      O'  «         «      ,  about  O"  negative. 

30.  Principle  of  Moments.      In  general,   a  single  force  of 
proper  magnitude  and  line  of  ac- 
tion can  balance  any  number  of 

forces.  That  single  force  is  called 
the  equilibrant  of  the  forces,  and 
the  single  force  that  would  balance 
the  equilibrant  is  called  the  result- 
ant of  the  forces.  Or,  otherwise 
stated,  the  resultant  of  any  num- 
ber of  forces  is  a  force  which  pro- 
duces the  same  effect.  It  can  be 
proved  that — The  algebraic  sum 
of  the  moments  of  any  number 
of  forces  with  respect  to  a  point, 
equals  the  moment  of  their  re-  Y\S.  1. 

sultant  about  that  point. 

*By  clockwise  direction  is  meant  that  in  which  the  hands  of   a  clock 
rotate;  and  by  counter-clockwise,  the  opposite  direction. 


18  STRENGTH  OF  MATERIALS 

This  is  a  useful  principle  and  is  called  "principle  of  moments." 

31.     All  the  forces  acting  upon  a  body  which  is  at  rest  are 

said  to  be  balanced  or  in  equilibrium.     "No  force  is  required  to 

balance  such  forces  and  hence  their  equilibrant  and  resultant  are 

zero. 

Since  their  resultant  is  zero,  the  algebraic  sum  of  the  mom- 


looolbs.        aooolbs.  aooolbs.  loooltos. 


A       r^  is 

5  v 

u         lC        -5  u 
Fig.  8. 

ents  of  any  number  of  forces  which  are  balanced  or  in  equilib- 
.  rium  equals  zero. 

This  is  known  as  the  principle  of  moments  for  forces  in 
equilibrium;  for  brevity  we  shall  call  it  also  "the  principle  of 
moments." 

The  principle  is  easily  verified  in  a  simple  case.  Thus,  let 
AB  (Fig.  8)  be  a  beam  resting  on  supports  at  C  and  F.  It  is 
evident  from  the  symmetry  of  the  loading  that  each  reaction 
equals  one-half  of  the  whole  load,  that  is,  \  of  6,000=3,000 
pounds.  (We  neglect  the  weight  of  the  beam  for  simplicity.) 

With  respect  to  C,  for  example,  the  moments  of  the  forces 
are,  taking  them  in  order  from  the  left: 

—1,000  X   4  =  —  4,000  foot-pounds 

3,000  X    0=  0        " 

2,000  X   2=       4,000 

2,000  X  14  =      28,000 
—3,000  X  16  =  —  48,000        " 

1,000  X  20  =      20,000 

The  algebraic  sum  of  these  moments  is  seen  to  equal  zero. 
Again,  with  respect  to  B  the  moments  are: 

— 1,000  X  24  =  —  24,000  foot-pounds 

3,000  X  20  =  60,000 
— 2,000.  X  18  =  -— 36,000 
—  2,000  X  6  =  —  12,000 

3,000  X    4=      12,000 

1,000  X    0=  0 

The  sum  of  these  moments  also  equals  zero.     In   fact,  no  matter 


STRENGTH  OF  MATERIALS  19 

where  the  center  of  moments  is  taken,  it  will  be  found  in  this  and 
any  other  balanced  system  of  forces  that  the  algebraic  sum  of  their 
moments  equals  zero.  The  chief  .use  that  we  shall  make  of  this 
principle  is  in  finding  the  supporting  forces  of  loaded  beams. 

32.  Kinds  of  Beams.     A  cantilever  'beam  is  one  resting  on 
one  support  or  fixed  at  one  end,  as  in  a  wall,  the  other  end  being 
free. 

A  simple  beam  is  one  resting  on  two  supports. 

A  restrained  beam  is  one  fixed  at  both  ends;  a  beam  fixed  at 
one  end  and  resting  on  a  support  at  the  other  is  said  to  be  re- 
strained at  the  fixed  end  and  simply  supported  at  the  other. 

A  continuous  beam  is  one  resting  on  more  than  two  supports. 

33.  Determination  of  Reactions  on  Beams.  The  forces  which 
the  supports  exert  on  a  beam,  that  is,  the  "supporting  forces,"  are 
called  reactions.     We  shall  deal  chiefly  with  simple  beams.     The 
reaction  on  a  cantilever  beam  supported  at  one  point  evidently 
equals  the  total  load  on  the  beam. 

When  the  loads  on  a  horizontal  beam  are  all  vertical  (and 

looolbs.  2000  Ibs.         aooolbs. 


L,       ,1 

51  

t.,1         J,          ^1        J 

K  —  1  * 

-    1     - 

'E 

T  B 

'R! 

C 

O                                                        i- 

D 
6»  * 

J     ^. 

t*..  £-•*• 

Fig.  9. 

^     - 

^  i 

this  is  the  usual  case),  the  supporting  forces  are  also  vertical  and 
the  sum  of  the  reactions  equals  the  sum  of  the  loads.  This  prin- 
ciple is  sometimes  useful  in  determining  reactions,  but  in  the  case 
of  simple  beams  the  principle  of  moments  is  sufficient.  The  gen- 
eral method  of  determining  reactions  is  as  follows: 

1.  Write  out  two  equations  of  moments  for  all  the  forces 
(loads  and  reactions)  acting  on  the  beam  with  origins  of  moments 
at  the  supports. 

2.  Solve  the  equations  for  the  reactions. 

3.  As  a  check,  try  if  the  sum  of  the  reactions  equals  the 
sum  of  the  loads. 

Examples.  1.  Fig.  9  represents  a  beam  supported  at  its 
ends  and  sustaining  three  loads.  We  wish  to  find  the  reactions 
due  to  these  loads. 


20  STRENGTH  OF  MATERIALS 

Let  the  reactions  be  denoted  by  Rt  and  R2  as  shown;  then 
the  moment  equations  are: 
For  origin  at  A, 

1,000  X 1  +  2,000  X  6  +  3,000  X  8— R2  X 10  =  0. 
For  origin  at  E, 

2100lbS.  3600  llOS. 


La,_ 6. 1 

I  [B JC     . 


ID 


Fig.  10. 

Ri  x  10—1,000  X  9—2,000  X  4—3,000  X  2  =  0. 
The  first  equation  reduces  to 

10  R2  =  1,000+12,000+24,000  =  37,000;  or 
R2=  3,700  pounds. 
The  second  equation  reduces  to 

10  E1=  9,000+8,000+6,000  =  23,000;  or 
B1==  2,300  pounds. 

The  sum  of  the  loads  is  6,000  pounds  and  the  sum  of  the  reactions 
is  the  same;  hence  the  computation  is  correct. 

2.  Fig.  10  represents  a  beam  supported  at  B  and  D  (that  is, 
it  has  overhanging  ends)  and  sustaining  three  loads  as  shown.  We 
wish  to  determine  the  reactions  due  to  the  loads. 

Let  Rj  and  R2  denote  the  reactions  as  shown ;  then  the  moment 
equations  are: 
For  origin  at  B, 

-2,100x2+0+3,600x6— R2X 14+ 1,600x18  =  0. 
For  origin  at  D, 

-2,100x16+^x14—3,600x8  +  0+1,600x4  =  0. 
The  first  equation  reduces  to 

14  Ea=  -4,200+21,600  +  28,800  =  46,200;  or 
R2  =  3,300  pounds. 
The  second  equation  reduces  to 

14  R1==  33,600+28,800-6,400  =  56,000;  or 
E1=  4,000  pounds. 

The  sum  of  the  loads  equals  7,300  pounds  and  the  sum  of  the 
reactions  is  the  same;  hence  the  computation  checks. 

3.  What  are  the  total   reactions  in  example  1  if  the  beam 
weighs  400  pounds  ? 


STRENGTH  OF  MATERIALS  21 

(1.)  Since  we  already  know  the  reactions  due  to  the  loads 
(2,300  and  3,700  pounds  at  the  left  and  right  ends  respectively 
(see  illustration  1  above),  we  need  only  to  compute  the  reactions 
due  to  the  weight  of  the  beam  and  add.  Evidently  the  reactions 
due  to  the  weight  equal  200  pounds  each;  hence  the 

left  reaction  =2,300+200—2,500  pounds,  and  the 
-right     «        =  3,700+ 200— 3,900       «       . 

(2.)  Or,  we  might  compute  the  reactions  due  to  the  loads 
and  weight  of  the  beam  together  and  directly.  In  figuring  the 
moment  due  to  the  weight  of  the  beam,  we  imagine  the  weight 
as  concentrated  at  the  middle  of  the  beam ;  then  its  moments  with 
respect  to  the  left  and  right  supports  are  (400  X  5)  and — (400  X  5) 
respectively.  The  moment  equations  for  origins  at  A  and  E  are 
like  those  of  illustration  1  except  that  they  contain  one  more 
term,  the  moment  due  to  the  weight ;  thus  they  are  respectively : 

1,000x1  +  2,000x6  +  3,000x8— R2X 10 +400X5-0, 

E!  X  10—1,000  X  9—2,000  X  4—3,000  X  2—400  X  5  -0. 
The  first  one  reduces  to 

10  E2=  39,000,  or  R,  =  3,900  pounds; 
and  the  second  to 

10  .E1=  25,000,  or  E1==  2,500  pounds. 

4.  What  are  the  total  reactions  in  example  2  if  the  beam 
weighs  42  pounds  per  foot  ? 

As  in  example  3,  we  might  compute  the  reactions  due  to  the 
weight  and  then  add  them  to  the  corresponding  reactions  due  to 
the  loads  (already  found  in  example  2),  but  we  shall  determine 
the  total  reactions  due  to  load  and  weight  directly. 

The  beam  being  20  feet  long,  its  weight  is  42x20,  or  840 
pounds.  Since  the  middle  of  the  beam  is  8  feet  from  the  left  and 
6  feet  from  the  right  support,  the  moments  of  the  weight  with 
to  the  left  and  right  supports  are  respectively: 

840X8  =  6,720,  and— 840x6  —  —5,040  foot-pounds. 
The  moment  equations   for  all  the   forces  applied  to   the    beam 
for  origins  at  B  and  D  are  like  those  in  example  2,  with  an  addi- 
tional term,  the  moment  of  the  weight;  "they  are  respectively: 
— 2,100x2  +  0+3,600x6— R2  X 14+ 1,600  X 18 +  6,720  =  0, 
—2,100  X 16 + Rj  X  14—3,600  X  8  +  0 + 1,600  X  4—5,040  =  0. 


22  STRENGTH  OF  MATERIALS 

The  first  equation  reduces  to 

14  R2=52,920,  or  R2=3,780  pounds, 
and  the  second  to 

14  R^  61,040,  or  R^  4,360  pounds. 

The  sum  of  the  loads  and  weight  of  beam  is  8,140  pounds; 
and  since  the  sum  of  the  reactions  is  the  same,  the  computation 
checks. 

EXAMPLES  FOR  PRACTICE. 

1.     AB  (Fig.  11)  represents  a  simple  beam  supported  at  its 
ends.     Compute  the  reactions,  neglecting  the  weight  of  the  beam, 

.        j  Right  reaction  =  1,443.75  pounds. 
''  J  Left  reaction     =  1,556.25  pounds. 

eoolbs.  900 Ibs.  soolbs.  looolbs. 


k— 2'— * 4- 


IB 


Fig.  11. 

2.  Solve  example  1  taking  into  account  the  weight  of  the 
beam,  which  suppose  to  be  400  pounds. 

.         (  Right  reaction  =  1,643.75  pounds. 
|  Left  reaction     =  1,756.25  pounds. 

3.  Fig.  12  represents  a  simple  beam  weighing  800  pounds 
supported  at  A  and  B,   and    sustaining  three  loads    as    shown. 
"What  are  the  reactions  ? 

.        j  Right  reaction  =  2,014.28  pounds. 
'  (  Left  reaction     =  4,785.72  pounds. 

2000  Ibs.  1000  Ibs.  3000  Ibs. 

—  •'  —     t         »'—      t 
* 


Fig.  12. 

4.  Suppose  that  in  example  3  the  beam  also  sustains  a  uni- 
formly distributed  load  (as  a  floor)  over  its  entire  length,  of  500 
pounds  per  foot.  Compute  the  reactions  due  to  all  the  loads  and 
the  weight  of  the  beam. 

.        j  Right  reaction  =    4,871.43  pounds. 
QS*  (  Left  reaction     =±=  11,928.57  pounds. 


STRENGTH  OF  MATERIALS  23 

EXTERNAL  SHEAR  AND  BENDING  MOMENT. 

On  almost  every  cross-section  of  a  loaded  beam  there  are 
three  kinds  of  stress,  namely  tension,  compression  and  shear.  The 
first  two  are  often  called  fibre  stresses  because  they  act  along  the 
real  fibres  of  a  wooden  beam  or  the  imaginary  ones  of  which  we 
may  suppose  iron  and  steel  beams  composed.  Before  taking  up 
the  subject  of  these  stresses  in  beams  it  is  desirable  to  study  certain 
quantities  relating  to  the  loads,  and  on  which  the  stresses  in  a 
beam  depend.  These  quantities  are  called  external  shear  and 
bending  moment,  and  will  now  be  discussed. 

34.  External  Shear.     By  external  shear  at  (or  for)  any  sec- 
tion of  a  loaded  beam  is  meant  the  algebraic  sum  of  all  the  loads 
(including  weight  of  beam)   and  reactions  on  either  side  of  the 
section.      This  sum  is  called  external  shear  because,  as  is  shown 
later,  it  equals  the  shearing  stress  (internal)  at  the  section.       For 
brevity,  we  shall  often  say  simply  "shear"  when  external  shear  is 
meant. 

35,  Rule  of  Signs.     In  computing  external  shears,  it  is  cus- 
tomary to  give  the  plus  sign  to  the  reactions  and  the  minus  sign 
to  the  loads.      But  in  order  to  get  the  same  sign  for  the  external 
shear  whether  computed  from  the  right  or  left,  we  change  the  sign 
of  the  sum  when  computed  from  the  loads  and  reactions  to  the 
right.     Thus  for  section  a  of  the  beam  in  Fig.  8  the  algebraic  sum  is, 
when  computed  from  the  left, 

-1,000  +  3,000=  +2,000  pounds; 
and  when  computed  from  the  right, 

-1,000+3,000-2,000-2,000  =  -2,000  pounds. 
The  external  shear  at  section  a  is  +2,000  pounds. 

Again,  for  section  b  the  algebraic  sum  is, 
when  computed  from  the  left, 

-1,000  +  3,000-2,000-2,000  +  3,000  =  +  1,000  pounds; 
and  when  computed  from  the  right,  -1,000  pounds. 

The  external  shear  at  the  section  is  + 1,000  pounds. 

It  is  usually  convenient  to  compute  the  shear  at  a  section 
from  the  forces  to  the  right  or  left  according  as  there  are  fewer 
forces  (loads  and  reactions)  on  the  right  or  left  sides  of  the 
section. 


24  STRENGTH  OF  MATERIALS 

36.  Units  for  Shears.      It  is  customary  to  express  external 
shears   in  pounds,  but  any  other  unit  for  expressing  force  and 
weight  (as  the  ton)  may  be  used. 

37.  Notation.     We  shall  .use  Y  to  stand  for  external  shear  at 
any  section,  and  the   shear  at  a  particular  section  will  be  denoted 
by  that  letter  subscripted;   thus  Y1?  Y2,  etc.,  stand  for  the  shears 
at  sections  one,  two,  etc.,  feet  from  the  left  end  of  a  beam. 

The  shear  has  different  values  just  to  the  left  and  right  of  a 
support  or  concentrated  load.  We  shall  denote  such  values  by  Y' 
and  Y";  thus  Y5'  and  Y5"  denote  the  values  of  the  shear  at  sec- 
tions a  little  less  and  a  little  more  than  5  feet  from  the  left  end 
respectively. 

Examples.  1.  Compute  the  shears  for  sections  one  foot 
apart  in  the  beam  represented  in  Fig.  9,  neglecting  the  weight  of 
the  beam.  (The  right  and  left  reactions  are  3,700  and  2,300 
pounds  respectively;  see  example  1,  Art.  33.) 

All  the  following  values  of  the  shear  are  computed  from  the 
left.  The  shear  just  to  the  right  of  the  left  support  is  denoted  by 
Y0",  and  Y0"  =  2,300  pounds.  The  shear  just  to  the  left  of  B  is 
denoted  by  Y/,  and  since  the  only  force  to  the  left  of  the  section 
is  the  left  reaction,  Y/  =  2,300  pounds.  The  shear  just  to  the 
right  of  B  is  denoted  by  Y/',  and  since  the  only  forces  to  the  left 
of  this  section  are  the  left  reaction  and  the  1,000-pound  load, 
Y/'  =  2,300  - 1,000  =  1,300  pounds.  To  the  left  oj  all  sections 
between  B  and  C,  there  are  but  two  forces,  the  left  reaction  and 
the  1,000-pound  load;  hence  the  shear  at  any  of  those  sections 
equals  2,300-1,000^1,300  pounds,  or 

Y2  =  Y3  =  Y4  =  Y5  =  Y6'=  1,300  pounds. 

The  shear  just  to  the  right  of  0  is  denoted  by  Y6";  and  since  the 
forces  to  the  left  of  that  section  are  the  left  reaction  and  the 
1,000-  and  2,000-pound  loads, 

Y6"  ==  2,300  - 1,000  -  2,00l  «  -  700  pounds. 

Without  further  explanation,  thb  student  should  understand 
that 

Y7    =  +  2,300  -  1,000  -  2,000  ===  -  700  pounds, 

Y;  =-700, 

Y8"  =  +  2,300  -  1,000  -  2,000  -  3,000  =  -  3,700, 

Y9  =  V10'=- 3,700, 

Y10"  =  +  2.300  - 1,000  -  2,000  -  3,000  +  3,700  =  0 


STRENGTH  OF  MATERIALS  25 

2.  A  simple  beam  10  feet  long,  and  supported  at  each  end, 
weighs  400  pounds,  and  bears  a  uniformly  distributed  load  of 
1,600  pounds.     Compute  the  shears  for  sections  two  feet  apart. 

Evidently  each  reaction  equals  one-half  the  sum  of  the  load 
and  weight  of  the  beam,  that  is,  \  (1,600+400)  =1,000  pounds. 
To  the  left  of  a  section  2  feet  from  the  left  end,  the  forces  acting 
on  the  beam  consist  of  the  left  reaction,  the  load  on  that  part  of 
the  beam,  and  the  weight  of  that  part  ;  then  since  the  load  and 
weight  of  the  beam  per  foot  equal  200  pounds, 

Y2=  1,000-200  X  2  =  600  pounds. 

To  the  left  of  a  section  four  feet  from  the  left  end,  the  forces 
are  the  left  reaction,  the  load  on  that  part  of  the  beam,  and  the 
weight ;  hence 

V \=  1,000-200  X  4  =  200  pounds. 

Without  further  explanation,  the  student  should  see  that 
Y6   =  1,000-200  X   6  =-200  pounds, 
Y8    =  1,000-200  X   8  =  -600  pounds, 
V10'  =  1,000-200  X 10  =  -1,000  pounds, 
V10"=  1,000-200x10+1,000  =  0. 

3.  Compute  the  values  of  the  shear  in  example  1,  taking 
into  account  the  weight  of  the  beam  (400  pounds).     (The  right 
and  left  reactions  are  then  3,900  and  2,500  pounds  respectively; 
see  example  3,  Art.  33.) 

We  proceed  just  as  in  example  1,  except  that  in  each  compu- 
tation we  include  the  weight  of  the  beam  to  the  left  of  the  section 
(or  to  the  right  when  computing  from  forces  to  the  right).  The 
weight  of  the  beam  being  40  pounds  per  foot,  then  (computing 
from  the  left) 

Vo"=+ 2,500  pounds, 

V/  =+2,500-40  =+2,460, 

V/'  =+2,500-40-1,000= +  1,460, 

V2   =+2,500-1,000-40x2  =+1,420, 

Y3    =  f  2,500-1,000-40  X  3  =  + 1,380, 

V.4   =+ 2,500-1,000-40  X  4  =  + 1,340, 

Y5   =+2,500-1,000-40x5  =  +1,300, 

V6'  =+ 2,500-1,000-40  X  6  =  + 1,260, 

Y  6"  =+ 2,500-1,000-40  X  6-2,000  =  -740, 

V7   =+2,500-1,000-2,000-40  X  7  =  -780, 


26  STRENGTH  OF  MATERIALS 

V8'  =  +  2,500-1,000-2,000-40  X  8  =  -820, 
V8"  =  +  2,500-1,000-2,000-40x8-3,000  =-3,820, 
Yg    =  +  2,500-1,000-2,000-3,000-40  X  9  =  -3,860, 
V'io  =  +  2,500-1,000-2,000-3,000-40  X 10  =  -3,900, 
Y"10=  +  2,500-1,000-2,000-3,000-40x10  +  3,900=0. 

Computing  from  the  right,  we  find,  as  before,  that 

V7   =-(3,900-3,000-40  X  3)  =-780  pounds, 
Y/  =_(35900-3,000-40X2)=-820, 
V8"  =-(3,900-40  X  2)^-3,820, 
etc.,    etc. 

EXAMPLES  FOR  PRACTICE. 

1.  Compute  the  values  of  the  shear  for  sections  of  the  beam 
represented  in  Fig.  10,  neglecting  the  weight  of  the  beam.     (The 
right  and  left  reactions  are  3,300  and  4,000  pounds  respectively; 
see  example  2,  Art.  33.) 

rV,    =V2'=- 2,100  pounds, 

Ans  J  Y*"  =V.=V4=VB=V6=VT=V'8=  + 1,900, 

'  1  V8"  =Y9=Y10=Y11=Y12=Y13=YU=Y15=Y'16=-1,700, 
L  V"16  =Y17=Y18-Y19=Y'20=  + 1,600. 

2.  Solve  the  preceding  example,  taking  into  accoun-t   the 
weight  of  the  beam,  42  pounds  per  foot.      (The  right  and  left 
reactions  are  3,780  and  4,360  pounds  respectively;  see  example  4, 
Art.  33.) 

Y0"  =  -  2,100  Ibs.   Y7   =  +  1,966  Ibs.   Yu  =- 1,928  Ibs. 

Y1    =-2,142  Y8'  =  +  1,924          Y15  =- 1,970 

Y2'  =  -  2,184  Y8"  =  - 1,676  Y16'  =  -  2,012 

Y2"  =  +  2,176  Y9   =-1,718  V16"= +1,768 

Y3  =  +  2,134  Y10  =  -  1,760  Y17  =  + 1,726 

y4   =  +  2,092  Yn  =-1,802  YI8  =  +  1,684 

Y5   =  +  2,050  Y12=- 1,844  Y19  =  +  1,642 

ye   =  +  2,008  Y13  =  -  1,886  VM'  =  + 1,600 

3.  Compute  the  values  of  the  shear  at  sections  one  foot  apart 
in  the  beam  of  Fig."  11,  neglecting  the  weight.      (The  right  and 
left  reactions  are  1,444  and  1,556  pounds  respectively;  see  example 
1,  Art.  33.) 


Ans. 


STRENGTH  OP  MATERIALS  27 


V0"  =V1=V/=  +  1,556  pounds, 
V,"  =¥,=¥,=¥,=¥;=  +956, 
V6"  =V/=  +  56, 
V,"  =V8=V9=V10=V11=V12=V13'=-444, 
[  V13"=V14=V15=V16'=-1,444. 

4.  Compute  the  vertical  shear  at  sections  one  foot  apart  ID 
the  beam  of  Fig.  12,  taking  into  account  the  weight  of  the  beam, 
800  pounds,  and  a  distributed  load  of  500  pounds  per  foot.  (The 
right  and  left  reactions  are  4,870  and  11,930  pounds  respectively; 
see  examples  3  and  4,  Art.  33.) 

Y0  =  0          Y7  =  +  6,150  Ibs.  Y15  =+    830  Ibs, 

Vx'  =  -  540  Ibs.  V8'  =  +5,610  Y16  =  +  290 
VY'=-  2,540  V8"  =+4,610  Y17'  =  -  250 
Y2  =  _  3,080  Y9  =+4,070  Y17"  =  -3,250 


Ans.  -< 


Y3  =  -  3,620         Y10  =  +  3,530         Y18  =  -3,790 


Y4  =  -  4,160  Yn  =  +2,990  Y19  =  -  4,330 

Y5  =  -  4,700  Y12  =  +2,450  Y20'  =  -  4,870 

Y6'  =  -  5,240  Y13  =  + 1,910  Y20"=  0 

Y6"=+ 6,690  VM=  + 1,370 

38.  Shear  Diagrams.  The  way  in  which  the  external  shear 
varies  from  section  to  section  in  a  beam  can  be  well  represented 
by  means  of  a  diagram  called  a  shear  diagram.  To  construct 
such  a  diagram  for  any  loaded  beam, 

1.  Lay   off  a  line  equal  (by  some  scale)   to  the  length  of 
the  beam,  and  mark  the  positions  of  the  supports  and  the  loads. 
(This  is  called  a  "base-line.") 

2.  Draw  a  line  such  that  the  distance  of  any  point  of  it 
from  the  base  equals  (by  some  scale)  the  shear  at  the  correspond- 
ing section  of  the  beam,  and  so  that  the  line  is  above  the  base 
where  the  shear  is  positive,  and  below  it  where  negative.     (This  is 
called  a  shear  line,  and  the  distance  from  a  point  of  it  to  the 
base  is  called  the  "ordinate"  from  the  base  to  the  shear  line  at 
that  point.) 

We  shall  explain  these  diagrams  further  by  means  of  illus- 
trative examples. 

Examples.  1.  It  is  required  to  construct  the  shear  diagram 
for  the  beam  represented  in  Fig.  13,  a  (a  copy  of  Fig.  9). 


28  STRENGTH  OF  MATERIALS 

Lay  off  A'E'  (Fig.  13,  &)  to  represent  the  beam,  and  mark  the 
positions  of  the. loads  B',  C'  and  D'.  In  example  1,  Art.  37,  we 
computed  the  values  of  the  shear  at  sections  one  foot  apart;  hence 
we  lay  off  ordinates  at  points  on  A'E'  one  foot  apart,  to  represent 
those  shears. 

Use  a  scale  of  4,000  pounds  to  one  inch.  Since  the  shear  for 
any  section  in  AB  is  2,300  pounds,  we  draw  a  line  ab  parallel 
to  the  base 0.575  inch (2,300 ^-4,000)  therefrom;  this  is  the  shear 
line  for  the  portion  AB.  Since  the  shear  for  any  section  in  BC 
equals  1,300  pounds,  we  draw  a  line  Vc  parallel  to  the  base  and 


looolbs. 


2000  Ibs.   3000  Ibs. 


ra~T~"             o               —  *  —  *  —  »  — 

2  M 

JB                             Tc           D 
i      i                                            i 
*.    b 

b1 

111  I  I  i    r    \, 

E' 

A       B'                             c1  Illllllllillllllllil 

c1              d 
ScdJe:  in-  4-000  Itos. 

d1     , 

'a. 


Fig.  13. 

0.325 inch (1,300-f- 4,000)  therefrom;  this  is  the  shear  line  for  the 
portion  BC.  Since  the  shear  for  any  section  in  CD  is  -700 
pounds,  we  draw  a  line  cd  below  the  base  and  0.175  inch 
(700-^-4,000)  therefrom;  this  is  the  shear  line  for  the  portion 
CD.  Since  the  shear  for  any  section  in  DE  equals  -3,700  Ibs.,  we 
draw  a  liued'e  below  the  base  and  0.925  inch  (3,700^-4,000)  there- 
from; this  is  the  shear  line  for  the  portion  DE.  Fig.  13,  £,  is  the 
required  shear  diagram. 

2.  It  is  required  to  construct  the  shear  diagram  for  the 
beam  of  Fig.  14,  a  (a  copy  of  Fig.  9),  taking  into  account  the 
weight  of  the  beam,  400  pounds. 

The  values  of  the  shear  for  sections  one  foot  apart  were  com- 
puted in  example  3,  Art.  37,  so  we  have  only  to  erect  ordinates  at 
the  various  noints  on  a  base  line  A'E'  (Tier.  14.  l>\  eaual  to  those 


STRENGTH  OF  MATERIALS 


29 


values.  We  shall  use  the  same  scale  as  in  the  preceding  illustra- 
tion, 4,000  pounds  to  an  inch.  Then  the  lengths  of  the  ordinates 
corresponding  to  the  values  of  the  shear  (see  example  3,  Art.  37} 
are  respectively: 

2,500-^4,000=0.625  inch 

2,460-^4,000=0.615     " 

1,460-^4,000=0.365     « 

etc.  etc. 

Laying  these  ordinates  off  from  the  base  (upwards  or  downwards 
according  as  they  correspond  to  positive  or  negative  shears),  we 
get  ob,  J'tf,  c'd,  and  d'e  as  the  shear  lines. 

iooolbs.  aooolfos.      3000  Ibs. 


,     , 

, 

c        *°        :c         2.'        ii 

r-1     - 

\ 

i         i 
i         i 

B 
b 

1 

c          ID 

. 

I 

i 

ii 

i 

i 

c          ! 
!D' 

El 


3.  It  is  required  to  construct  the  shear  diagram  for  the 
cantilever  beam  represented  in  Fig.  15,  a,  neglecting  the  weight 
of  the  beam. 

The  value  of  the  shear  for  any  section  in  AB  is  —  500  pounds ; 
for  any  section  in  BC,  -1,500  pounds;  and  for  any  section  in 
CD,  -  3,500  pounds.  Hence  the  shear  lines  are  ab,  b'c,  c'd.  The 
scale  being  5,000  pounds  to  an  inch, 

A'«  =     500-^5,000  =  0.1  inch, 

B'£'  =  1,500-^5,000  =  0.3     « 

CV  =  3,500-^5,000  =  0.7     « 

The  shear  lines  are  all  below  the  base  because  all  the  values  of  the 
shear  are  negative. 


30 


STRENGTH  OF  MATERIALS 


4.  Suppose  that  the  cantilever  of  the  preceding  illustration 
sustains  also  a  uniform  load  of  200  pounds  per  foot  (see  Fig.  16,  a). 
Construct  a  shear  diagram. 


2000  Ibs. 


First,  we  compute  the  values  of  the  shear  at  several  sections. 
Thus      V0"  ==-  500  pounds, 

V,    =-500 -200= -700, 

Y2'  =_500-200x2^-900, 
Y2"  =_  500  -  200  X  2  -  1,000^-1,900, 
Y3   =-  500  -  1,000  -  200  X  3=-.2,100, 
Y4    ==-  500  -  1,000  -  200x4^-2,300, 
Y5'  =-500  -  1,000  -  200x5^-2,500, 
Y5"  =-  500  -  1,000  -  200X5  -  2,000—4,500, 
V6    =-  500  -  1,000  -  2,000  -  200x6^-4,700, 
Y7    =_  500  -  1,000  -  2,000  -  200  X  7^-4,900, 
Y8    =-  500  -  1,000  -  2,000  -  200  X  8=-5,100, 
Y9    =  -500  -  1,000  -  2,000  -  200x9^-5,300. 
The  values,  being  negative,  should  be  plotted  downward.     To  a 
scale  of  5,000  pounds  to  the  inch  they  give  the  shear  lines  #5,  b'c, 
c<d  (Fig.  16,  I). 

EXAMPLES  FOR  PRACTICE. 

1.  Construct  a  shear  diagram  for  the  beam  represented  in 
Fig.  10,  neglecting  the  weight  of  the  beam  (see  example  1,  Art.  37). 

2.  Construct  the  shear  diagram  for  the  beam  represented  in 
Fig.  11,  neglecting   the  weight    of  the    beam    (see    example    3, 
Art  37). 


STRENGTH  OF  MATERIALS 


31 


3.  Construct  the  shear  diagram  for  the  beam  of  Fig.  12 
when  it  sustains,  in  addition  to  the  loads  represented,  its  own 
weight,  800  pounds,  and  a  uniform  load  of  500  pounds  per  foot 
(see  example  4,  Art.  37). 

4.  Figs,  a,  cases  1  and  2,  Table  B,  represent  two  cantilever 
beams,  the  first  bearing  a  concentrated  load  P  at  the  free  end, 
and  the  second  a  uniform  load  W.     Figs,  b  are  the  corresponding 
shear  diagrams.     Take  P  and  W  equal  to  1,000  pounds,  and  satisfy 
yourself  that  the  diagrams  are  correct. 

and  satisfy  yourself  that  the  diagrams  are  correct. 

5.  Figs.  «,    cases  3   and  4,  same   table,   represent   simple 
beams  supported  at  their  ends,  the  first  bearing  a  concentrated 

500  Ibs.  looolbs.       zooolbs. 


£-„                      0                       , 

I 

lA        IB 
i           i 
j           i 

!A          'B' 

C                     D* 
C1                     D'| 

u 

Sca-le  i"= sooo Ibs.  d 

Fig.  16. 

load  P  at  the  middle,  and  the  second  a  uniform  load  W.  Figs. 
b  are  the  corresponding  shear  diagrams.  Take  P  and  W  equal 
to  1,000  pounds,  and  satisfy  yourself  that  they  are  correct. 

39.  Maximum  Shear.  It  is  sometimes  desirable  to  know 
the  greatest  or  maximum  value  of  the  shear  in  a  given  case.  This 
value  can  always  be  found  with  certainty  by  constructing  the  shear 
diagram,  from  which  the  maximum  value  of  the  shear  is  evident  at 
a  glance.  In  any  case  it  can  most  readily  be  computed  if  one 
knows  the  section  for  which  the  shear  is  a  maximum.  The  stu- 
dent should  examine  all  the  shear  diagrams  in  the  preceding 
articles  and  those  that  he  has  drawn,  and  see  that 

1.  In  cantilevers  fixed  in  a  wall,  the  maximum  shear 
occurs  at  the  wall. 


32  STRENGTH  OF  MATERIALS 

2.  In  simple  beams,  the  maximum  shear  occurs  at  a  sec- 
tion next  to  one  of  the  supports. 

By  the  use  of  these  propositions  one  can  determine  the  value 
of  the  maximum  shear  without  constructing  the  whole  shear 
diagram.  Thus,  it  is  easily  seen  (referring  to  the  diagrams,  page 
53)  that  for  a 

Cantilever,  end  load  P,  maximum  shear=P 

«         ,  uniform  load  W,         "  "     =W 

Simple  beam,  middle  load  P,        "  "     =±P 

"  «    ,  uniform  «   W,        «  "     =±W 

40.  Bending  floment.     By  bending  moment  at  (or  for)  a 
section  of  a  loaded  beam,  is  meant  the  algebraic  sum  of  the  mo- 
ments of  all  the  loads   (including  weight  of  beam)  and  reactions 
to  the  left  or  right  of  the  section  with  respect  to  any  point  in  the 
section. 

41.  Rule  of  Signs.     We  follow  the  rule  of  signs  previously 
stated  (Art.  29)  that  the  moment  of  a  force  which  tends  to  pro- 
duce clockwise  rotation  is  plus,  and  that  of  a  force  which  tends  to 
produce  counter-clockwise  rotation  is  minus;  but  in  order  to  get 
the  same  sign  for  the  bending  moment  whether  computed  from 
the  right  of  left,  we  change  the  sign  of  the  sum  of  the  moments 
when  computed  from  the  loads  and  reactions  on  the  right.     Thus . 
for  section  a,  Fig.   8,  the  algebraic  sums  of  the  moments  of  the 
forces  are : 

when  computed  from  tne  left, 

-1,000  X  5  +  3,000  X  1—2,000  foot-pounds ; 
and  when  computed  from  the  right, 

1,000  X  19-3,000  X 15  +  2,000  X 13  +  2,000  X 1 =  +  2,000  foot- 
pounds. 
The  bending  moment  at  section  a  is  -2,000  foot-pounds. 

Again,  for  section  o,  the  algebraic  sums  of  the  moments  of  the 
forces  are: 
when  computed  from  the  left, 

-1,000  X  22  +  3,000x18-2,000  X  16-2,000  X  4 + 3,000  X  2= 

-2,000  foot-pounds; 
and  when  computed  from  the  right, 

1,000x2= +  2,000  foot-pounds. 
The  bending  moment  at  the  section  is  -2.000  foot-pounds. 


STRENGTH  OF  MATERIALS  33 

It  is  usually  convenient  to  compute  the  bending  moment  for 
a  section  from  the  forces  to  the  right  or  left  according  as  there 
are  fewer  forces  (loads  and  reactions)  on  the  right  or  left  side 
of  the  section. 

42.  Units.     It  is  customary  to  express  bending  moments -in 
inch -pounds,  but  often   the  foot-pound  unit  is  more  convenient. 
To  reduce  foot-pounds  to  inch-pounds,  multiply  ~by  twelve. 

43.  Notation.     We  shall  use  M  to  denote  bending  moment  at 
any  section,  and  the  bending  moment  at  a  particular  section  will 
be  denoted  by  that  letter  subscripted;  thus  M1?   M2,    etc.,    denote 
values  of  the  bending  moment  for  sections  one,  two,  etc.,  feet 
from  the  left  end  of  the  beam. 

Examples.  1.  Compute  the  bending  moments  for  sections 
one  foot  apart  in  the  beam  represented  in  Fig.  9,  neglecting  the 
weight  of  the  beam.  (The  right  and  left  reactions  are  3,700  and 
2,300  pounds  respectively.  See  example  1,  Art.  33.) 

Since  there  are  no  forces  acting  on  the  beam  to  the  left  of  the 
left  support,  M0— 0.  To  the  left  of  the  section  one  foot  from  the 
left  end  there  is  but  one  force,  the  left  reaction,  and  its  arm  is  one 
foot;  hence  M,  =  +  2,300x1=2,300  foot-pounds.  To  the  left  of 
a  section  two  feet  from  the  left  end  there  are  two  forces,  2,300  and 
1,000  pounds,  and  their  arms  are  2  feet  and  1  foot  respectively; 
hence  M2=  +  2,300x2-1,000x1  =3,600  foot-pounds.  At  the 
left  of  all  sections  between  B  and  C  there  are  only  twro  forces, 
2,300  and  1,000  pounds;  hence 

M3=  +  2,300  X  3-1,000  X  2= +4,900  foot-pounds, 
M4=  +  2,300  X  4-1,000  X  3=  +  6,200        « 
M5=  +  2,300  X  5-1,000  X  4=  +  7,500        " 
M0=  +  2,300x6-1,000x5=  +  8,800     '   « 

To  the  right  of  a  section  seven  feet  from  the  left  end  there 
are  two  forces,  the  3,000-pound  load  and  the  right  reaction 
(3,700  pounds),  and  their  arms  with  respect  to  an  origin  in  that 
section  are  respectively  one  foot  and  three  feet;  hence 

M7= -(-3,700  X  3  +  3,000  x  1) =  +  8,100  foot-pounds. 

To  the  right  of  any  section  between  E  and  D  there  is  only  one 
force,  the  right  reaction ;  hence 


34  STRENGTH  OF  MATERIALS 

M8=-(-3,700  X  2)^7,400  foot-pounds, 
M9 =-(-3,700  X  1)=3,700 

Clearly  M10=0. 

2.  A  simple  beam  10  feet  long  and  supported  at  its  ends 
weighs  400  pounds,  and  bears  a  uniformly  distributed  load  of  1,600 
pounds.  Compute  the  bending  moments  for  sections  two  feet 
apart. 

Each  reaction  equals  one-half  the  whole  load,  that  is,  -|  of 
.(1,600+400)  =1,000  pounds,  and  the  load  per  foot  including 
weight  of  the  beam  is  200  pounds.  The  forces  acting  on  the 
beam  to  the  left  of  the  first  section,  two  feet  from  the  left  end,  are 
the  left  reaction  (1,000  pounds)  and  the  load  (including  weight) 
on  the  part  of  the  beam  to  the  left  of  the  section  (400  pounds). 
The  arm  of  the  reaction  is  2  feet  and  that  of  the  400 -pound  force 
is  1  foot  (the  distance  from  the  middle  of  the  400 -pound  load  to 
the  section).  Hence 

M2=  + 1,000  X  2-400  X 1  =  + 1,600  foot-pounds. 

The  forces  to  the  left  of  the  next  section,  4  feet  from  the  left 
end,  are  the  left  reaction  and  all  the  load  (including  weight  of 
beam)  to  the  left  (800  pounds).  The  arm  of  the  reaction  is  4  feet, 
and  that  of  the  800-pound  force  is  2  feet;  hence 

M4=  + 1,000  X  4-800  X  2 =  +  2,400  foot-pounds. 
Without  further  explanation  the  student  should  see  that 

M6=  + 1,000  X  6-1,200  X  3=  +  2,400  foot-pounds, 
M8=  + 1,000  X  8-1,600  X  4= + 1,600 

Evidently  M0=M10=0. 

8.  Compute  the  values  oj;  the  bending  moment  in  example 
1,  taking  into  account  the  weight  of  the  beam,  400  pounds.  (The 
right  and  left  reactions  are  respectively  3,900  and  2,500  pounds; 
see  example  3,  Art.  33.) 

We  proceed  as  in  example  1,  except  that  the  moment 
of  the  weight  of  the  beam  to  the  left  of  each  section  (or  to 
the  right  when  computing  from  forces  to  the  right)  must  be 
included  in  the  respective  moment  equations.  Thus,  computing 
from  the  left, 


STRENGTH  OF  MATERIALS 


35 


M0  -0 

Mt  =  +  2,500  X 1-40  X  j=  +  2,480  foot-pounds, 

M2  =  +  2,500  X  2-1,000  X 1-80  X 1=  +  3,920, 

M3  =  +  2,500  X  3-1,000  X  2-120  X  1£=  +  5,320, 

M4  =  +  2,500  X  4-1,000  X  3-160  X  2=+  6,680, 

M5  =  +  2,500  X  5-1,000  X  4-200  X  2±=  +  8,000, 

M,  =+ 2,500  X  6-1,000  X  5-240  X  3=  +  9,280. 

Computing  from  the  right, 

M7  =-(-3,900x3+3,OOOxl  +  120x 

M8  =-(-3,900x2  +  80xl)  =  +  7,720, 

M9  =-(-3,900xl+40x4)=+3,880, 


M10  =  0. 


EXAMPLES  FOR  PRACTICE. 


1.  Compute  the  values  of  the  bending  moment  for  sections 
one  foot  apart,  beginning  one  foot  from  the  left  end  of  the 
beam  represented  in  Fig.  10,  neglecting  the  weight  of  the  beam. 
(The  right  and  left  reactions  are  3,300  and  4,000  pounds  respec- 
tively; see  example  2,  Art.  33.) 

M,=  -  2,100  M6  =  +  3,400  Mn=  +  2,100  M16=-6,400 
M2=  -  4,200  M7  =  +  5,300  M12=  +  400  M17=-4,800 
M3=  -  2,300  M.  =  +  7,200  M13=  -  1,300  M18=-3,200 


Ans. 
(in  foot- 
pounds) 


M4=  -     400  M,  =  +  5,500  M14=  -  3,000  M19=-l,600 
M5=  +  1,500  M10=  +  3,800  M15=  -  4,700  M20=         0 

2.  Solve  the   preceding  example,  taking  into  account   the 
weight  of  the  beam,  42  pounds  per  foot.     (The  right  and  left 
reactions  are  3,780  and  4,360  pounds  respectively;  see  example  4, 
Art.  33.) 

'Mx=  -  2,121  M6  =+4,084  Mn=  +  2,799  M16==  -  6,736 

Ans.       M2=  -  4,284  M7  =  +  6,071  M12=+    976  M17=  -  4,989 

(in  foot.  J  M3=  -  2,129  M8  =  +  8,016  M13=  -     889  M18=  -  3,284 

pounds)     MI==  _       16  M9  =  +  6,319  Mu=  -  2,796  M19=  -  1,621 

\  M5=  +  2,055  M10=+  4,580  M15=  -  4,745  M20=  0 

3.  Compute  the  bending  moments    for  sections    one   foot 
apart,  of  the  beam  represented  in  Fig.  11,  neglecting  the  weight. 
(The  right  and  left  reactions  are  1,444  and  1,556  pounds  respect- 
ively;  see  example  1,  Art.  33.) 


36 


STRENGTH  OF  MATERIALS 


pounds) 


,=  +  l,556  M5=  +  5,980  M9  =+6,104  M,,=  +4,328 


Ma=+<M>68  M7=+6,992  Mn=  +  5,216  MI6=  +  1,440 
M4=  +  5,024  M8=  +  6,548  M12=  +  4,772  M16=  0 


4  Compute  the  bending  moments  at  sections  one  foot  apart 
in  the  beam  of  Fig.  12,  taking  into  account  the  weight  of  the  beam, 
800  pounds,  and  a  uniform  load  of  500  pounds  per  foot.  (The 
right  and  left  reactions  are  4,870  and  11,930  pounds  respectively; 
see  Exs.  3  and  4,  Art.  33.) 

[X^-     270  M6  =-19,720  Mn=+  3,980  M16=12,180 

Ans.       Ma=-  3,080  M7  =  -13,300  M12=+  6,700  M17=  12,200 

(in  foot  J  M8=-  6,430  M8  =-  7,420  M13=+  8,880  M18=  8,680 

pounds)     M4=  -10,320  M9  =-  3,080  M14=  +  10,520  M19=  4,620 

L  M6=  -14,750  Mlo=  +     720  M1B=  +  11,620  M20=         0 

44.  Moment  Diagrams.  The  way  in  which  the  bending 
moment  varies  from  section  to  section  in  a  loaded  beam  can  be 
well  represented  by  means  of  a  diagram  called  a  moment  diagram. 
To  construct  such  a  diagram  for  any  loaded  beam, 


1000  Ibs. 


sooolbs.      3000  Itas. 
2' 


ScaJie:i  «  looooft.-lbs 
Fig.  17. 

1.  Lay  off  a  base-line    just   as   for  a  shear  diagram   (see 
Art.  38). 

2.  Draw  a  line  such  that  the  distance  from  any  point  of  it 
to  the  base-line  equals  (by  some  scale)  the  value  of  the  bending 
moment  at  the  corresponding  section  of  the  beam,  and  so  that  the 
line  is  above  the  base  where  the  bending  moment  is  positive  and 
below  it  where  it  is  negative.    (This  line  is  called  a  "moment 
line."} 


STRENGTH  OF  MATERIALS  37 

Examples.  1.  It  is  required  to  construct  a  moment  dia- 
gram for  the  beam  of  Fig.  17,  a  (a  copy  of  Fig.  9),  loaded  as 
there  shown. 

Layoff  A'E'  (Fig.  17,  5)  as  a  base.  In  example  1,  Art.  43, 
we  computed  the  values  of  the  bending  moment  for  sections  one 
foot  apart,  so  we  erect  ordinates  at  points  of  A'E'  one  foot  apart, 
to  represent  the  bending  moments. 

We  shall  use  a  scale  of  10,000  foot-pounds  to  the  inch;  then 
<he  ordinates  (see  example  1,  Art.  43,  for  values  of  M)  will  be: 
One  foot  from  left  end,  2,300-^-10,000  =  0.23  inch, 
Two  feet     «       «      "     3,600-^-10,000  =  0.36     " 
Three  "      «       «      «     4,900-^10,000  =  0.49     " 
Four   "      «       «      "     6,200-^-10,000  =  0.62     " 
etc.,    etc. 


•r 

looolbs. 
I                         5' 

2000  Ibs. 

1 

30OO  Ib3. 
«i           1           -i           J 

:, 

|                         3 

T 

-       T      * 

t 

IB 

p 

!D 

1 

i 
1 

.^TTTTrfflTn'niTnii 

TTfTTTTTTTTnUi 

B1  C1  D1  E' 

ScoJe:  i"-  looooft.-lbs. 
Fig.  18. 

Laying  these  ordinates  off,  and  joining  their  ends  in  succession, 
we  get  the  line  A'&<%£E',  which  is  the  bending  moment  line. 
Fig.  17,  b,  is  the  moment  diagram. 

2.  It  is  required  to  construct  the  moment  diagram  for  the 
beam,  Fig.  18,  a  (a  copy  of  Fig.  9),  taking  into  account  the  weight 
of  -the  beam,  400  pounds. 

The  values  of  the  bending  moment  for  sections  one  foot  apart 
were  computed  in  example  3,  Art.  43.  So  we  have  only  to  lay  off 
ordinates  equal  to  those  values,  one  foot  apart,  on  the  base  A'E; 
(Fig.  18,  b). 

To  a  scale  of  10,000  foot-pounds  to  the  inch  the  ordinates 
(see  example  3,  Art.  43,  for  values  of  M)  are: 


38 


STRENGTH  OF  MATERIALS 


At  left  end,  0 

One  foot  from  left  end,  2,480-^10,000=0.248  inch 
Two  feet     «       "      «      3,920-^-10,000=0.392    « 
Three  «      "       «      «      5,320-f- 10,000=0.532    « 
Four    "      «       «      «      6,680-^-10,000=0.668    « 
Laying  these  ordinates  off  at  the  proper  points,  we  get 
as  the  moment  line. 

3.  It  is  required  to  construct  the  moment  diagram  for  the 
cantilever  beam  represented  in  Fig.  19,  #,  neglecting  the  weight 
of  the  beam.  The  bending  moment  at  B  equals 

-500x2=-l,000  foot-pounds; 
atC, 

-500  X  5-1,000  X  3=-5,500 ; 
and  at  D, 

-500  X  9-l,OOOX  7-2,000  X  4=-19,500. 

500  Ibs.    looolbs. 


Sca,1e:i"*  sooooft.-ltas 


Fig.  19. 

Using  a  sca/e  of  20,000  foot-pounds  to  one  inch,  the  ordinates 
in  the  bending  moment  diagram  are: 

AtB,    1,000-5-20,000=0.05  inch, 

«  C,    5,500-^20,000=0.275  « 

«  D,  19,500^-20,000=0.975  « 

Hence  we  lay  these  ordinates  off,  and  downward  because  the  bend- 
ing moments  are  negative,  thus  fixing  the  points  5,  c  and  d.  The 
bending  moment  at  A  is  zero;  hence  the  moment  line  connects  A 
&,  c  and  d.  Further,  the  portions  A£,  be  and  cd  are  straight,  as 
can  be  shown  by  computing  values  of  the  bending  moment  for 
sections  in  AB,  BC  and  CD,  and  laying  off  the  corresponding 
ordinates  in  the  moment  diagram. 


STRENGTH  OF  MATERIALS 


39 


4.  Suppose  that  the  cantilever  of  the  preceding  illustration 
sustains  also  a  uniform  load  of  100  pounds  per  foot  (see  Fig.  20,  a). 
Construct  a  moment  diagram. 

First,  we  compute  the  values  of  the  bending  moment  at  sev- 
eral  sections;  thus, 

Mx=-500  X 1-100  X  J=-550  foot-pounda, 
Ma=-500x  2-200  XI =-1,200, 
Ms=-500  X  3-1,000  x  1-300  X 1  J=-2,950, 
M4=-500  X  4-1,000  X  2-400  X  2=-4,800, 
M5=-500  X  5-1,000  X  3-500  X  2J= -6,750, 
M6=-500  X  6-1,000  X  4-2,000  X 1-600  X  3=-10,800, 
M°=:-500  X  7-1,000  X  5-2,000  X  2-700  X  3£=-14,950, 
M8=-500  X  8-1,000  X  6-2,000  X  3-800  X  4=-19,200, 
1L=-500  X  9-1,000  X  7-2,000  X  4-900  X  4£=-23,550. 


500  Ibs.  1000 Ibs. 

3' 


2000  Ibs. 


ScaJe: 


Fig.  20. 

These  values  all  being  negative,  the  ordinates  are  all  laid  off 
downwards.  To  a  scale  of  20,000  foot-pounds  to  one  inch,  they 
fix  the  moment  line  A.'bcd. 

EXAflPLES  FOR  PRACTICE. 

1.  Construct  a  moment  diagram  for  the  beam  represented  in 
Fig.  10,  neglecting   the   weight  of  the  beam.    (See  example  1, 
\rt.  43). 

2.  Construct  a  moment  diagram   for  the  beam  represented 
in  Fig.  11,  neglecting  the  weight  of  the  beam.      (See  example  3, 
Art.  43). 

3.  Construct  the  moment  diagram  for  the  beam  of  Fig.  12 


40  STRENGTH  OF  MATERIALS 

when  it  sustains,  in  addition  to  the  loads  represented  and  its  own 
weight  (800  pounds),  a  uniform  load  of  500  pounds  per  foot. 
(See  example  4,  Art.  43.) 

4.  Figs,  a,  cases  1  and  2,  page  53,  represent  two  cantilever 
beams,  the  first  bearing  a  load  P  at  the  free  end,  and  the  second 
a   uniform    load    W.      Figs,  c    are  the  corresponding   moment 
diagrams.     Take  P  and  "W  equal  to  1,000  pounds,  and  I  equal  to 
10  feet,  and  satisfy  yourself  that  the  diagrams  are  correct. 

5.  Figs.  #,  cases  3  and  4,  page  53,  represent  simple  beams 
on  end  supports,  the  first  bearing  a  middle  load  P,  and  the  other  a 
uniform  load  W.      Figs,  c  are  the  corresponding  moment  dia- 
grams.    Take  P  and  "W  equal  to  1,000  pounds,  and  I  equal  to 
10  feet,  and  satisfy  yourself  that  the  diagrams  are  correct. 

45.  Maximum  Bending  Moment.  It  is  sometimes  desirable 
to  know  the  greatest  or  maximum  value  of  the  bending  moment 
in  a  given  case.  This  value  can  always  be  found  with  certainty 
by  constructing  the  moment  diagram,  from  which  the  maximum 
value  of  the  bending  moment  is  evident  at  a  glance.  But  in  any 
case,  it  can  be  most  readily  computed  if  one  knows  the  section  for 
which  the  bending  moment  is  greatest.  If  the  student  will  com- 
pare the  corresponding  shear  and  moment  diagrams  which  have 
been  constructed  in  foregoing  articles  (Figs.  13  and  17,  14  and 
18,  15  and  19,  16  and  20),  and  those  which  he  has  drawn,  he  will 
see  that — The  maximum  bending  moment  in  a  beam  occurs 
where  the  shear  changes  sign. 

By  the  help  of  the  foregoing  principle  we  can  readily  com- 
pute  the  maximum  moment  in  a  given  case.  We  have  only  to 
construct  the  shear  line,  and  observe  from  it  where  the  shear 
changes  sign;  then  compute  the  bending  moment  for  that  section. 
If  a  simple  beam  has  one  or  more  overhanging  ends,  then  the  shear 
changes  sign  more  than  once — twice  if  there  is  one  overhanging 
end,  and  three  times  if  two.  In  such  cases  we  compute  the 
bending  moment  for  each  section  where  the  shear  changes  sign; 
the  largest  of  the  values  of  these  bending  moments  is  the  maxi- 
mum for  the  beam. 

The  section  of  maximum  bending  moment  in  a  cantilever 
fixed  at  one  end  (as  when  built  int3  a  wall)  is  always  at  the  wall. 


STRENGTH  OF  MATERIALS  41 

Thus,  without  reference  to  the  moment  diagrams,  it  is  readily  seen 
that, 

for  a  cantilever  whose  length  is  Z, 

with  an  end    load  P,  the  maximum  moment  is      PZ, 
"    a  uniform  «     W,  "  "  "        "  J  "W7. 

Also  by  the  principle,  it  is  seen  that, 

for  a  beam  whose  length  is  Z,  on  end  supports, 

with  a  middle  load  P,  the  maximum  moment  is  ^  PZ, 
«    uniform     "     W,  «  "  "         "  J-  WZ. 

46.  Table  of  Maximum   Shears,   Moments,  etc.     Table  B 
on  page  53    shows  the    shear   and  moment  diagrams   for   eight 
simple  cases  of  beams.    The  first  two  cases  are  built-in  cantilevers; 
the  next  four,  simple  beams  on  end  supports;  and  the  last  two, 
restrained  beams  built  in  walls  at  each    end.      In   each   case    I 
denotes  the  length. 

CENTER  OF  GRAVITY  AND  HOMENT  OF  INERTIA. 

It  will  be  shown  later  that  the  strength  of  a  beam  depends 
partly  on  the  form  of  its  cross-section.  The  following  discussion 
relates  principally  to  cross -sections  of  beams,  and  the  results 
reached  (like  shear  and  bending  moment)  will  be  made  use  of 
later  in  the  subject  of  strength  of  beams. 

47.  Center  of  Gravity  of  an  Area.     The  student  probably 
knows  what  is  meant  by,  and  how  to  find,  the  center  of  gravity  of 
any  flat  disk,  as  a  piece  of  tin.     Probably  his  way  is  to  balance 
the  piece  of  tin  on  a  pencil  point,  the  point  of  the  tin  at  which  it  so 
balances  being  the  center  of  gravity.    (Really  it  is  midway  between 
the  surfaces  of  the  tin  and  over  the  balancing  point.)     The  center 
of  gravity  of  the  piece  of  tin,  is  also  that  point  of  it  through  which 
the  resultant  force  of  gravity  on  the  tin  (that  is,  the  weight  of  the 
piece)  acts. 

By  "center  of  gravity"  of  a  plane  area  of  any  shape  we  mean 
that  point  of  it  which  corresponds  to  the  center  of  gravity  of  a 
piece  of  tin  when  the  latter  is  cut  out  in  the  shape  of  the  area. 
The  center  of  gravity  of  a  quite  irregular  p,rea  can  be  found  most 
readily  by  balancing  a  piece  of  tin  or  stiff  paper  cut  in  the  shape 
of  the  area.  But  when  an  area  is  simple  in  shape,  or  consists  of 
parts  which  are  simple,  the  center  of  gravity  of  the  whole  can  be 


42  STRENGTH  OF  MATERIALS 

found  readily  by  computation,  and  such  a  method  will  now  be 
described. 

48.  Principle  of  floments  Applied  to  Areas.  Let  Fig.  21 
represent  a  piece  of  tin  which  has  been  divided  off  into  any  num- 
ber of  parts  in  any  way,  the  weight  of  the  whole  being  "W,  and 
that  of  the  parts  "W,,  W2,  W3,  etc.  Let  Ow  C2,  C8,  etc.,  be  the 
centers  of  gravity  of  the  parts,  C  that  of  the  whole,  and  <?n  <?2,  <?3, 
etc.,  and  c  the  distances  from  those  centers  of  gravity  respectively 
to  some  line  (L  L)  in  the  plane 
of  the  sheet  of  tin.  When  the 
tin  is  lying  in  a  horizontal  posi- 
tion, the  moment  of  the  weight 
of  the  entire  piece  about  L  L  is 
"We,  and  the  moments  of  the 
parts  are  W^,  W2<?2,  etc.  Since 
the  weight  of  the  whole  is  the  . 

resultant  of  the  weights  of  the 

o  Fig.  21. 

parts,  the  moment  of  the  weight 

of  the  whole  equals  the  sum  of  the  moments  of  the  weights  of  the 
parts;  that  is, 


Now  let  A:,  A2,  etc.  denote  the  areas  of  the  parts  of  the  pieces 
of  tin,  and  A  the  area  of  the  whole;  then  since  the  weights  are 
proportional  to  the  areas,  we  can  replace  the  "W's  in  the  preceding 
equation  by  corresponding  A's,  thus: 

Ac=A1<?1  +  A2<?2-fetc  .....  (4) 

If  we  call  the  product  of  an  area  and  the  distance  of  its 
center  of  gravity  from  some  line  in  its  plane,  the  "moment"  of  the 
area  with  respect  to  that  line,  then  the  preceding  equation  may  be 
stated  in  words  thus: 

The  moment  of  an  area  with  respect  to  any  line  equals  the 
algebraic  sum  of  the  moments  of  the  parts  of  the  area. 

If  all  the  centers  of  gravity  are  on  one  side  of  the  line  with 
respect  to  which  moments  are  taken,  then  all  the  moments  should  be 
given  the  plus  sign;  but  if  some  centers  of  gravity  are  on  one  side 
and  some  on  the  other  side  of  the  line,  then  the  moments  of  the 
areas  whose  centers  of  gravity  are  on  one  side  should  be  given  the 


STRENGTH  OF  MATERIALS  43 

same  sign,  and  the  moments  of  the  others  the  opposite  sign.  The 
foregoing  is  the  principle  of  moments  for  areas,  and  it  is  the  basis 
of  all  rules  for  finding  the  center  of  gravity  of  an  area. 

To  find  the  center  of  gravity  of  an  area  which  can  be  divided 
np  into  simple  parts,  we  write  the  principle  in  forms  of  equations 
for  two  different  lines  as  "axes  of  moments,"  and  then  solve  the 
equations  for  the  unknown  distances  of  the  center  of  gravity  of  the 
whole  from  the  two  lines.  We  explain  further  by  means  of  specific 
examples. 

Examples.  1.  It  is  required  to  find  the  center  of  gravity 
of  Fig.  22,  a,  the  width  being  uniformly  one  inch. 

The  area  can  be  divided  into  two  rectangles.     Let  Ct  and 


•C 


•C2 


1 


^  To 

Fig.  22. 

C2  be  the  centers  of  gravity  of  two  such  parts,  and  C  the  center  of 
gravity  of  the  whole.  Also  let  a  and  ~b  denote  the  distances  of  C 
from  the  two  lines  OL'  and  OL"  respectively. 

The  areas  of  the  parts  are  6  and  3  square  inches,  and  their 
arms  with  respect  to  OL'  are  4  inches  and  \  inch  respectively,  and 
with  respect  to  OL"  -J  inch  and  1^  inches.  Hence  the  equations  of 
moments  with  respect  to  OL'  and  OL"  (the  whole  area  being  9 
square  inches)  are: 


9x^=6x1+3x14=75. 

Hence,  a  —  25.5-^-9  —  2.83  inches, 

I  =    7.5-V-9  =  0.83      «     . 

2.     It  is  required  to  locate  the  center  of  gravity  of  Fig.  22,  b, 
the  width  being  uniformly  one  inch. 


44 


STRENGTH  OF  MATERIALS 


The  figure  can  be  divided  up  into  three  rectangles.  Let  Cj,  C2 
and  C3  be  the  centers  of  gravity  of  such  parts,  C  the  center  of 
gravity  of  the  whole;  and  let  a  denote  the  (unknown)  distance  of 
C  from  the  base.  The  areas  of  the  parts  are  4,  3  0  and  4  square 
inches,  and  their  "  arms  "  with  respect  to 
the  base  are  2,^  and  2  inches  respectively; 
hence  the  equation  of  moments  with  re- 
spect to  the  base  (the  en  tire  area  being  18 
square  inches)  is: 

18X0  =  4x2+10X-J+4x2  =  21. 
Hence,  a  —  21-r-18  —  1.17  inches. 

From  the  symmetry  of  the  area  it  is  plain 
that  the  center  of  gravity  is  midway  be- 
tween the  sides. 


EXAMPLE  FOR  PRACTICE. 

1.    Locate  the  center  of   gravity   of 
Fig.  23.  Fig.  23. 

Ans.     2.3  inches  above  the  base. 

49.  Center  of  Gravity  of  Built=up  Sections.  In  Fig.  24 
there  are  represented  cross-sections  of  various  kinds  of  rolled  steel, 
called  "shape  steel,"  which  is  used  extensively  in  steel  construction. 
Manufacturers  of  this  material  publish  "handbooks"  giving  full 
information  in  regard  thereto,  among  other  things,  the  position  of 
the  center  of  gravity  of  each  cross  section.  With  such  a  handbook 


1-foea.m 


Channel 


Angle 


Z-b^r 


Fig.  2L 

available,  it  is  therefore  not  necessary  actually  to  compute  the  posi- 
tion of  the  center  of  gravity  of  any  section,  as  we  did  in  the  pre- 
ceding article;  but  sometimes  several  shapes  are  riveted  together  to 


STRENGTH  OF  MATERIALS 


45 


make  a  "built-up"  section  (see  Fig.  25),  and  then  it  may  be  neces- 
sary to  compute  the  position  of  the  center  of  gravity  of  the  section. 
Example.     It  is  desired  to  locate  the  center  of  gravity  of  the 
section  of  a  built-up  beam  represented  in  Fig.  25.    The  beam  con- 


14"- 


CVJ 


Fig.  25. 

sists  of  two  channels  and  a  plate,  the  area  of  the  cross -section  of  a 

channel  being  6.03  square  inches. 

Evidently  the  center  of  gravity  of  each  channel  section  is  6 

inches,  and  that  of  the  plate  section  is  12^  inches,  from  the  bottom. 

Let  c  denote  the  dis- 
tance of  the  center  of 
gravity  of  the  whole 
section  from  the  bot- 
tom; then  since  the 
area  of  the  plate  section 
is  7  square  inches,  and 
that  of  the  whole  sec- 
tion is  19.06, 

19.06X^  =  6.03X6+ 
6.03X6  +  7  X12J  = 
158.11. 
Hence,  c=  158.11-^-19.06=8.30  inches,  (about). 

EXAMPLES  FOR  PRACTICE. 

1.     Locate  the  center  of  gravity  of  the  built-up  section  of 


* 


b 


26- 


46  STRENGTH  OF  MATERIALS 

Fig.  26,  #,  the  area  of  each  "angle"  being  5.06  square  inches,  and 
the  center  of  gravity  of  each  being  as  shown  in  Fig.  26,  b. 

Ans.     Distance  from  top,  3.08  inches. 

2.  Omit  the  left-hand  angle  in  Fig.  26,  0,  and  locate  the 
Center  of  gravity  of  the  remainder. 

.         (  Distance  from  top,  3.65  inches, 
'"  |          "  "      left  side,  1.19  inches. 

50.  floment  of  Inertia.  If  a  plane  area  be  divided  into  an 
infinite  number  of  infinitesimal  parts,  then  the  sum  of  the  prod- 
ucts obtained  by  multiplying  the  area  of  each  part  by  the  square 
of  its  distance  from  some  line  is  ealled  the  moment  of  inertia,  of  the 
area  with  respect  to  the  line.  The  line  to  which  the  distances  are 
measured  is  called  the  inertia-axis;  it  may  be  taken  anywhere  in 
the  plane  of  the  area.  In  the  subject  of  beams  (where  we  have 
sometimes  to  compute  the  moment  of  inertia  of  the  cross  -section 
of  a  beam),  the  inertia-axis  is  taken  through  the  center  of  gravity 
of  the  section  and  horizontal. 

An  approximate  value  of  the  moment  of  inertia  of  an  area 
can  be  obtained  by  dividing  the  area  into  small  parts  (not  infini- 
tesimal), and  adding  the  products  obtained  by  multiplying  the 
area  of  each  part  by  the  square  of  the  distance  from  its  center  to 
the  inertia-axis. 

Example.  If  the  rectangle  of  Fig.  27,  #,  is  divided  into  8 
parts  as  shown,  the  area  of  each  is  one  square  inch,  and  the  dis- 
tances from  the  axis  to  the  centers  of  gravity  of  the  parts  are  \ 
and  1^  inches.  For  the  four  parts  lying  nearest  the  axis  the 
product  (area  times  distance  squared)  is: 

lX(  -J)2=4;  and  for  the  other  parts  it  is 


Hence  the  approximate  value  of  the  moment  of  inertia  of  the  area 
with  respect  to  the  axis,  is 


If  the  area  is  divided  into  32  parts,  as  shown  in  Fig.  27,  J, 
the  area  of  each  part  is  J  square  inch.  For  the  eight  of  the  little 
squares  farthest  away  from  the  axis,  the  distance  from  their  centers 
of  gravity  to  the  axis  is  1J  inches;  for  the  next  eight  it  is  1J; 
for  the  next  eight  J;  and  for  the  remainder  J  inch.  Hence  an 


STRENGTH  OF  MATERIALS 


47 


approximate  value  of  the  moment  of  inertia  of  the  rectangle  with 
respect  to  the  axis  is : 


,2!L 


If  we  divide  the  rectangle  into  still  smaller  parts  and  form 
the  products 

(small  area)  X  (distance)2, 

and  add  the  products  just  as  we  have  done,  we  shall  get  a  larger 
answer  than  ,10J.  The  smaller  the  parts  into  which  the  rec- 
tangle is  divided,  the  larger  will  be  the  answer,  but  it  will  never 
be  larger  than  10§.  This  10§  is  the  sum  corresponding  to  a 

division  of  the  rectangle  into  an 
infinitely  large  number  of  parts 
(infinitely  small)  and  it  is  the 
exact  value  of  the  moment  of 
inertia  of  the  rectangle  with  re- 
spect to  the  axis  selected. 

There  are  short  methods   of 
computing  the  exact  values  of  the 
„.     9  moments  of  inertia  of  simple  fig- 

ures (rectangles,  circles,  etc.,), 

but  they  cannot  be  given  here  since  they  involve  the  use  of  difficult 
mathematics.  The  foregoing  method  to  obtain  approximate  val- 
ues of  moments  of  inertia  is  used  especially  when  the  area  is  quite 
irregular  in  shape,  but  it  is  given  here  to  explain  to  the  student 
the  meaning  of  the  moment  of  inertia  of  an  area.  He  should 
understand  now  that  the  moment  of  inertia  of  an  area  is  sim- 
ply a  name  for  such  sums  as  we  have  just  computed.  The  name 
is  not  a  fitting  one,  since  the  sum  has  nothing  whatever  to  do  with 
inertia.  It  was  first  used  in  this  connection  because  the  sum  is 
very  similar  to  certain  other  sums  which  had  previously  been 
called  moments  of  inertia. 

51.  Unit  of  Moment  of  Inertia.  The  product  (area  X  dis- 
tance2) is  really  the  product  of  four  lengths,  two  in  each  factor  ; 
and  since  a  moment  of  inertia  is  the  sum  of  such  products,  a 
moment  of  inertia  is  also  the  product  of  four  lengths.  Now  the 
product  of  two  lengths  is  an  area,  the  product  of  three  is  a  vol- 
ume, and  the  product  of  four  is  moment  of  inertia — unthinkable  in 


48  STRENGTH  OF  MATERIALS 

the  way  in  which  we  can  think  of  an  area  or  volume,  and  there- 
fore the  source  of  much  difficulty  to  the  student.  The  units  of 
these  quantities  (area,  volume,  and  moment  of  inertia)  are  respec- 
tively : 

the  square  inch,  square  foot,  etc., 
"  cubic  "  ,  cubic  "  "  , 
"  biquadratic  inch,  biquadratic  foot,  etc.; 

but  the  biquadratic  inch  is  almost  exclusively  used  in  this  connec- 
tion; that  is,  the  inch  is  used  to  compute 
values  of  moments  of  inertia,  as  in  the  pre- 
ceding illustration.  It  is  often  written 
thus:  Inches4, 
•i  52.  Moment  of  Inertia  of  a  Rectangle. 


J 

w  - 


Let  1}  denote  the  base  of  a  rectangle,  and  a 
Pig  28.  its  altitude;  then  by  higher  mathematics  it 

can  be  shown  that  the  moment  of  inertia 

of  the  rectangle  with  respect  to  a  line  through  its  center  of  gravity 
and  parallel  to  its  base,  is  y1^  W. 

Example.  Compute  the  value  of  the  moment  of  inertia  of 
a  rectangle  4x12  inches  with  respect  to  a  line  through  its  center 
of  gravity  and  parallel  to  the  long  side. 

Here  5=12,  and  a  =  4  inches  ;  hence  the  moment  of  inertia 
desired  equals 

TV(12X43)= 64  inches4. 

EXAHPLE  FOR  PRACTICE. 

1.  Compute  the  moment  of  inertia  of  a  rectangle  4x12 
inches  with  respect  to  a  line  through  its  center  of  gravity  and 
parallel  to  the  short  side.  Ans.  576  inches4. 

53.  Reduction  Formula.  In  the  previously  mentioned 
"handbooks"  there  can  be  found  tables  of  moments  of  inertia  of 
all  the  cross-sections  of  the  kinds  and  sizes  of  rolled  shapes  made. 
The  inertia-axes  in  those  tables  are  always  taken  through  the  cen- 
ter of  gravity  of  the  section,  and  usually  parallel  to  some  edge  of 
the  section.  Sometimes  it  is  necessary  to  compute  the  moment  of 
inertia  of  a  "rolled  section"  with  respect  to  some  other  axis,  and 
if  the  two  axes  (that  is,  the  one  given  in  the  tables,  and  the  other) 
are  parallel,  then  the  desired  moment  of  inertia  can  be  easily  com- 
puted from  the  one  given  in  the  tables  by  the  following  rule: 


STRENGTH  OF  MATERIALS 


49 


-co 


=2 


The  moment  of  inertia  of  an  area  with  respect  to  any  axis 
equals  the  moment  of  inertia  with  respect  to  a  parallel  axis 
through  the  center  of  gravity r,  plus  the  product  of  the  area  and 
the  square  of  the  distance  between  the  axes. 

Or.  if  I  denotes  the  moment  of  inertia  with  respect  to  any  axis; 
I0  the  moment  of  inertia  with  respect  to  a  parallel  axis  through 
the  center  of  gravity;  A  the  area;  and  d  the  (^.stance  between  the 
axes,  then 

I=Io+A^....  (5) 

Example.  It  is  required  to  compute  the  moment  of  inertia 
of  a  rectangle  2x8  inches  with  respect  to  a  line  parallel  to  the 
long  side  and  4  inches  from  the  center  of  gravity. 

Let  I  denote  the  moment  of  inertia  sought,  and  I0  the  moment 
of  inertia  of  the  rectangle  with  respect 
to  a  line  parallel  to  the  long  side  and 
through  the  center  of  gravity  (see  Fig. 
28).     Then 

I0=^5a8  (see  Art.  52);  and, 
since  £=8  inches  and  a=2  inches, 
Io=Tiir(8x23)=51L  biquadratic  inches. 

The  distance  between  the  two  inertia- 
axes  is  4  inches,  and  the  area  of  the 
rectangle  is  16  square  inches,  hence 
equation  5  becomes  Fi£'  29» 

I=5iL-f  16x42=261iL  biquadratic  inches. 

EXAMPLE  FOR  PRACTICE, 

1.  The  moment  of  inertia  of  an  "angle"  2-|x2X-J  inches 
(lengths  of  sides  and  width  respectively)  with  respect  to  a  line 
through  the  center  of  gravity  and  parallel  to  the  long  side,  is  0.64 
inches*.  The  area  of  the  section  is  2  square  inches,  and  the  dis- 
tance from  the  center  of  gravity  to  the  long  side  is  0.63  inches. 
(These  values  are  taken  from  a  "handbook".)  It  is  required  to 
compute  the  moment  of  inertia  of  the  section  with  respect  to  a 
line  parallel  to  the  long  side  and  4  inches  from  the  center  of 
gravity.  Ans.  32.64  inches*. 

54.  Moment  of  Inertia  of  Built-up  Sections.  As  before 
stated,  beams  are  sometimes  "built  up"  of  rolled  shapes  (angles, 


K 


50 


STRENGTH  OF  MATERIALS 


channels,  etc.).  The  moment  of  inertia  of  such  a  section  with 
respect  to  a  definite  axis  is  computed  by  adding  the  moments  of 
inertia  of  the  parts,  all  with  respect  to  that  same  axis.  This  is  the 
method  for  computing  the  moment  of  any  area  which  can  be 
divided  into  simple  parts. 

The  moment  of  inertia  of  an  area  which  may  be  regarded  as 
consisting  of  a  larger  area  minus  other  areas,  is  computed  by  sub- 
tracting from  the  moment  of  inertia  of  the  large  area  those  of  the 
"minus  areas." 

Examples.     1.  Compute  the  moment  of  inertia  of  the  built- 
up  section  represented  in  Fig.  30  (in  part  same  as  Fig.  25)  with 
respect  to  a  horizontal  axis 
passing  through  the  center 
of  gravity,  it  being  given 
that  the  moment  of  inertia 
of    each    channel    section 
with  respect  to  a  horizontal      *§" 
axis  through  its  center  of 
gravity   is    128.1   inches4, 
and    its    area  6.03  square 
inches. 

The  center  of  gravity  of 
the  whole  section  was  found 

in  the  example  of  Art.  49  to  be  8.30  inches  from  the  bottom  of 
the  section;  hence  the  distances  from  the  inertia-axis  to  the 
centers  of  gravity  of  the  channel  section  and  the  plate  are  2.30 
and  3.95  inches  respectively  (see  Fig.  30). 

The  moment  of  inertia  of  one  channel  section  with  respect  to 
the  axis  A  A  (see  equation  5,  Art.  53)  is: 

128.1  +  6.03  X2.302=160.00  inches4. 

The  moment  of  inertia  of  the  plate  section  (rectangle)  with  re- 
spect  to  the  line  a" a"  (see  Art.  52)  is: 

^  ba3=rV[14><  (i)3H°-15  inches4; 
and  with  respect  to  the  axis  A  A  (the  area  being  7  square  inches) 

it  is: 

0.15+ 7  X3.952=109.37  inches4. 

Therefore  the  moment  of  inertia  of  the  whole  section  with    re- 

spect  to  AA  is : 

2x160.00+109.37=429.37  inches*. 


STRENGTH  OF  MATERIALS 


51 


2.  It  is  required  to  compute  the  moment  of  inertia  of  the 
"  hollow  rectangle  "  of  Fig.  29  with  respect  to  a  line  through  the 
center  of  gravity  and  parallel  to  the  short  side. 

The  amount  of  inertia  of  the  large  rectangle  with  respect  to 
the  named  axis  (see  Art.  52)  is: 


B 


<pi.66"     a. 


B 

Fig.  31. 

and  the  moment  of  inertia  of  the  smaller  one  with  respect  to  the 
same  axis  is: 

TV(4X8°)  =  170§; 

hence  the  moment  of  inertia  of  the  hollow  section  with  respect 
to  the  axis  is: 

416§  -  170§  =-246  inches4. 
EXAMPLES  FOR  PRACTICE. 

1.  Compute    the   moment  of  inertia  of  the  section    repre- 
sented in  Fig.  31,  <z,  about  the  axis  AA,  it  being  3.08   inches 
from  the  top.     Given  also  that  the  area  of  one  angle  section  is 
5.06  square  inches,  its  center  of  gravity  C  (Fig.  31,  b)  1.66  inches 
from  the  top,  and  its  moment  of  inertia  with  respect  to  the  axis  aa 
17.68  inches4.  Ans.     145.8  inches*. 

2,  Compute  the  moment  of  inertia  of  the  section  of  Fig.  31,  a, 


52 


STRENGTH  OF  MATERIALS 


with  respect  to  the  axis  BB.  Given  that  distance  of  the  center 
of  gravity  of  one  angle  from  one  side  is  1.66  inches  (see  Fig.  31,5), 
and  its  moment  of  inertia  with  respect  to  bb  17.68  inches. 

Ans.  77.618  inches4. 

55.  Table  of  Centers  of  Gravity  and  floments  of  Inertia. 
Column  2  in  Table  A  below  gives  the  formula  for  moment  of 
inertia  with  respect  to  the  horizontal  line  through  the  center  of 
gravity.  The  numbers  in  the  third  column  are  explained  in  Art. 
62;  and  those  in  the  fourth,  in  Art.  80. 

TABLE  A. 

Moments  of  Inertia,  Section  Moduli,  and  Radii  of  Gyration. 

In  each  case  the  axis  is  horizontal  and  passes  through  the  center  of  gravity. 


Section. 


Moment  of 
Inertia. 


Section 
Modulus. 


Radius  of 
Gyration. 


a4 
12 


1/12 


ba3 
12 


12 


0.049d* 


0.049  (d4-d/) 


baf 
6 


6a 


0.098d3 


12 


0.098- 


l/d24-d,2 


STRENGTH  OF  BEAMS. 

56.  Kinds  of  Loads  Considered.  The  loads  that  are  applied 
to  a  horizontal  beam  are  usually  vertical,  but  sometimes  forces  are 
applied  otherwise  than  at  right  angles  to  beams.  Forces  acting  on 
beams  at  right  angles  are  called  transverse  forces ;  those  applied 


STRENGTH  OF  MATERIALS 


53 


TABLE  B. 

Shear  Diagrams  (b)  and  Moment  Diagrams  (c)  for  Eight  Different  Cases  (a). 
Also   Values  of  Maximum   Shear  (V),   Bending   floment  (M),  and   Deflection   (d). 


'.  f 


V=P,  M=P1,  d=P!3H-3EI. 


d=W!3-H8EI. 


JW-anfform 


,   M=MP1,   d=  P18-S-48EI. 


=J^  W,  M=%W1,  d=5Wl3-j-384EI. 


,     M=Pab-l. 


V=P,  M=Pa,  d=Pa(312-4a2)-?-24EI. 


d=P!3-f-192EI. 


54 


STRENGTH  OF  MATERIALS 


parallel  to  a  beam  are  .called  longitudinal  forces ;  and  others  are 
called  inclined  forces.  For  the  present  we  deal  only  with  beams 
subjected  to  transverse  forces  (loads  and  reactions). 

57.  Neutral  Surface,  Neutral  Line,  and  Neutral  Axis.  When 
a  beam  is  loaded  it  may  be  wholly  convex  up  (concave  down),  as  a 
cantilever;  wholly  convex  down  (concave  up),  as  a  simple  beam 
on  end  supports;  or  partly  convex  up  and  partly  convex  down,  as 
a  simple  beam  with  overhanging  ends,  a  restrained  beam,  or  a  con- 


Fig.  32. 

tinuous  beam.  Two  vertical  parallel  lines  drawn  close  together  on 
the  side  of  a  beam  before  it  is  loaded  will  not  be  parallel  after  it 
is  loaded  and  bent.  If  they  are  on  a  convex-down  portion  of  a 
beam,  they  will  be  closer  at  the  top  and  farther  apart  below  than 
when  drawrn  (Fig.  32&),  and  if  they  are  on  a  convex-up  portion, 
they  will  be  closer  below  and  farther  apart  above  than  when  drawn 
(Fig.  325). 

The  "  fibres  "  on  the  convex  side  of  a  beam  are  stretched  and 
therefore  under  tension,  while  those  on  the  concave  side  are  short- 
ened and  therefore  under  compression.  Obviously  there  must  be 
some  intermediate  fibres  which  are  neither  stretched  nor  shortened, 
i.  e.,  under  neither  tension  nor  compression.  These  make  up 
a  sheet  of  fibres  and  define  a  surface  in  the  beam,  which  surface  is 
called  the  neutral  surface  of  the  beam.  The  intersection  of  the 
neutral  surface  with  either  side  of  the  beam  is  called  the  neutral 
line,  and  its  intersection  with  any  cross-section  of  the  beam  is 
called  the  neutral  axis  of  that  section.  Thus,  if  ab  is  a  fibre  that 
has  been  neither  lengthened  nor  shortened  with  the  bending  of  the 
beam,  then  nn  is  a  portion  of  the  neutral  line  of  the  beam;  and, 
if  Fig.  32<?  be  taken  to  represent  a  cross-section  of  the  beam,  NN 
is  the  neutral  axis  of  the  section. 

It  can  be  proved  that  the  neutral  axis  of  any  cross-section  of 


STRENGTH  OF  MATERIALS 


55 


a  loaded  bearn  passes  through  the  center  of  gravity  of  that  section, 
provided  that  all  the  forces  applied  to  the  beam  are  transverse,  and 
that  the  tensile  and  compressive  stresses  at  the  cross-section  are 
all  within  the  elastic  limit  of  the  material  of  the  beam. 

58.  Kinds-  of  Stress  at  a  Cross=section  of  a  Beam.  It  has 
already  been  explained  in  the  preceding  article  that  there  are  ten- 
sile and  compressive  stresses  in  a  beam,  and  that  the  tensions  are 
on  the  convex  side  of  the  beam  and  the  compressions  on  the  con- 
cave  (see  Fig.  33).  The  forces  T  and  C  are  exerted  upon  the 
portion  of  the  beam  represented  by  the  adjoining  portion  to  the 


or 


i  T 


or 


Fig.  33. 

right  (not  shown).  These,  the  student  is  reminded,  are  often  called 
fibre  stresses. 

Besides  the  fibre  stresses  there  is,  in  general,  a  shearing  stress 
at  every  cross-section  of  a  beam.  This  may  be  proved  as  follows: 

Fig.  34  represents  a  simple  beam  on  end  supports  which  has 
actually  been  cut  into  two  parts  as  shown.  The  twro  parts  can 
maintain  loads  when  in  a  horizontal  position,  if  forces  are  applied 
at  the  cut  ends  equivalent  to  the  forces  that  would  act  there  if  the 
beam  were  not  cut.  Evidently  in  the  solid  beam  there  are  at  the 
section  a  compression  above  and  a  tension  below,  and  such  forces 
can  be  applied  in  the  cut  beam  by  means  of  a  short  block  C  and  a 
chain  or  cord  T,  as  shown.  The  block  furnishes  the  compressive 
forces  and  the  chain  the  tensile  forces.  At  first  sight  it  appears  as 
if  the  beam  would  stand  up  under  its  load  after  the  block  and 
chain  have  been  put  into  place.  Except  in  certain  cases*,  how- 
ever,  it  would  not  remain  in  a  horizontal  position,  as  would  the 

*  When  the  external  shear  for  the  section  is  zero. 


56 


STRENGTH  OF  MATERIALS 


solid  beam.  This  shows  that  the  forces  exerted  by  the  block  and 
chain  (horizontal  compression  and  tension  )  are  not  equivalent  to 
the  actual  stresses  in  the  solid  beam.  What  is  needed  is  a  vertical 
force  at  each  cut  end. 

Suppose  that  Rj  is  less  than  L1  +  L2  + weight  of  A,  i.  e.,  that 
the  external  shear  for  the  section  is  negative;  then,  if  vertical' pulls 
be  applied  at  the  cut  ends,  upward  on  A  and  downward  on  B,  the 
beam  will  stand  under  its  load  and  in  a  horizontal  position,  just  as 
a  solid  beam.  These  pulls  can  be  supplied,  in  the  model  of  the 
beam,  by  means  of  a  cord  S  tied  to  two  brackets  fastened  on  A  and 


Fig.  34. 


///ft///  RI 

K 


w 


Fig.  35. 


B,  as   shown.     In  the  solid    beam    the  two  parts  act  upon  each 
other  directly,  and  the  vertical  forces  are  shearing  stresses,  since 
they  act  in  the  plane  of  the  surfaces  to  which  they  are  applied. 

59.  Relation  Between  the  Stress  at  a  Section  and  the  Loads 
and  Reactions  on  Either  Side  of  It.  Let  Fig.  35  represent  the 
portion  of  a  beam  on  the  left  of  a  section ;  and  let  Rx  denote  the 
left  reaction;  Lj  and  L2  the  loads;  W  the  weight  of  the  left  part; 

C,  T,  and  S  the  compression,  tension,  and  shear  respectively  which 
the  right  part  exerts  upon  the  left. 

Since  the  part  of  the  beam  here  represented  is  at  rest,  all  the 
forces  exerted  upon  it  are  balanced;  and  when  a  number  of  hori- 
zontal and  vertical  forces  are  balanced,  then 

1.  The  algebraic  sum  of  the  horizontal  forces  equals  zero. 

2.  "  «  "      "    "    vertical          "  "        " 

3.  "  "  "      "    "    moments  of  all  the  forces  with  respect  to 
any  point  equals  zero. 

To  satisfy  condition  1,  since  the  tension  and  compression  are 

the  only  horizontal  forces,  the  tension  must  equal  the  compression. 

To  satisfy  condition  2,  S  (the  internal  shear)  must  equal  the 


STRENGTH  OF  MATERIALS  57 

algebraic  sum  of  all  the  other  vertical  forces  on  the  portion,  that 
is,  must  equal  the  external  shear  for  the  section;  also,  S  must  act 
up  or  down  according  as  the  external  shear  is  negative  or  positive. 
In  other  words,  briefly  expressed,  the  internal  and  external  shears 
at  a  section  are  equal  and  opposite* 

To  satisfy  condition  3,  the  algebraic  sum  of  the  moments  of 
the  fibre  stresses  about  the  neutral  axis  must  be  equal  to  the  sum 
of  the  moments  of  all  the  other  forces  acting  on  the  portion  about 
the  same  line,  and  the  signs  of  those  sums  must  be  opposite.  (The 
moment  of  the  shear  about  the  neutral  axis  is  zero.)  Now,  the 
sum  of  the  moments  of  the  loads  and  reactions  is  called  the  bend- 
ing moment  at  the  section,  and  if  we  use  the  term  resisting  mo- 
ment to  signify  the  sum  of  the  moments  of  the  fibre  stresses  (ten- 
sions and  compressions  )  about  the  neutral  axis,  then  we  may  say 
briefly  that  the  resisting  and  the  bending  moments  at  a  section  are 
equal,  and  the  two  moments  are  opposite  in  sign. 

60.  The  Fibre  Stress.  As  before  stated,  the  fibre  stress  is 
not  a  uniform  one,  that  is,  it  is  not  uniformly  distributed  over  the 
section  on  which  it  acts.  At  any  section,  the  compression  is  most 
"  intense  "  (or  the  unit-compressive  stress  is  greatest)  on  the  con- 
cave side;  the  tension  is  most  intense  (or  the  unit-tensile  stress  is 
greatest)  on  the  convex  side;  and  the  unit-compressive  and  unit- 
tensile  stresses  decrease  toward  the  neutral  axis,  at  which  place  the 
unit-fibre  stress  is  zero. 

If  the  fibre  stresses  are  within  the  elastic  limit,  then  the  two 
straight  lines  on  the  side  of  a  beam  referred  to  in  Art.  57  will  still 
be  straight  after  the  beam  is  bent;  hence  the  elongations  and  short- 
enings of  the  fibres  vary  directly  as  their  distance  from  the  neutral 
axis.  Since  the  stresses  (if  within  the  elastic  limit)  and  deforma- 
tions in  a  given  material  are  proportional,  the  unit-fibre  stress 
varies  as  the  distance  from  the  neutral  axis. 

Let  Fig.  36&  represent  a  portion  of  a  bent  beam,  365  its  cross- 
section,  nn  the  neutral  line,  and  NN  the  neutral  axis.  The  way 
in  which  the  unit-fibre  stress  on  the  section  varies  can  be  rep- 
resented graphically  as  follows:  Lay  off  ac,  by  some  scale,  to 
represent  the  unit-fibre  stress  in  the  top  fibre,  and  join  c  and  n, 
extending  the  line  to  the  lower  side  of  the  beam ;  also  make  be*  equal 
to  be?  and  draw  ncf.  Then  the  arrows  represent  the  unit-fibre 
stresses,  for  their  ^ngthsvary  as  their  distances  from  the  neutral  axis. 


58 


STRENGTH  OF  MATERIALS 


6i.  Value  of  the  Resisting  Moment.  If  S  denotes  the  unit- 
fibre  stress  in  the  fibre  farthest  from  the  neutral  axis  (the  greatest 
unit-fibre  stress  on  the  cross-section),  and  c  the  distance  from  the 
neutral  axis  to  the  remotest  fibre,  while  Sn  S2,  S3,  etc.,  denote  the 
unk-fibre  stresses  at  points  whose  distances  from  the  neutral  axis 
are,  respectively,  y,,  y^  y^  etc.  (see  Fig.  36 £),  then 

Q 

S   :  S,   ::  c  :  y,;  or  S,  =  -~yr 

G 

Q  QJ 

Also,  S2  =  ~  y2 ;  S3  =—yu  etc. 


Let 


a 


etc.,  be  the  areas  of  the  cross-sections  of  the  fibres 
S    c 


dw  10 

Fig.  36. 

whose  distances  from  the  neutral  axis  are,  respectively,  yl9  y^  y^ 
etc.     Then  the  stresses  on  those  fibres  are,  respectively, 
Sj  al9  S2  ^2,  S3  «3,  etc.; 

S  S  S 

Or,  y\&V      ^2^2J      — ^3^3?    etC* 

c  c  c 

The  arms  of  these  forces  or  stresses  with  respect  to  the  neutrai  axis 
are,  respectively,  y^  y^  y^  etc.;  hence  their  moments  are 

S       „    S      „     S 

etc., 


and  the  sum  of  the  moments  (that  is,  the  resisting  moment)  is 

Q  Ql  Q 

— «i  y!  +-7^2  y\  + etc-  =  v^1  y*  ~4~a*$  + etc>) 

Now  ^  y\  +  a2  y\  -f  etc.  is  the  sum  of  the  products  obtained  by 
multiplying  each  infinitesimal  part  of  the  area  of  the  cross-section 
by  the  square  of  its  distance  from  the  neutral  axis; hence,  it  is  the 
moment  of  inertia  of  the  cross-section  with  respect  to  the  neutral 
axis.  If  this  moment  is  denoted  by  I,  then  the  value  of  the  resist- 

f.    SI 
ing  moment  is  — o 


STRENGTH  OF  MATERIALS, 

PART  II. 


STRENGTH  OF  BEANS—  (Concluded). 

62.  First    Beam    Formula.     As    shown    in    the   preceding 
article,  the  resisting  and  bending  moments  for  any  section  of  a 
beam  are  equal;  hence 

?  =  M,  (6) 

all  the  symbols  referring  to  the  same  section.  This  is  the  most 
important  formula  relating  to  beams,  and  will  be  called  the  "  first 
beam  formula." 

The  ratio  I  H-  c  is  now  quite  generally  called  the  section 
modulus.  Observe  that  for  a  given  beam  it  depends  only  on  the 
dimensions  of  the  cross-section,  and  not  on  the  material  or  any- 
thing else.  Since  I  is  the  product  of  four  lengths  (see  article  51), 
I  -r-  c  is  the  product  of  three;  and  hence  a  section  modulus  can  be 
expressed  in  units  of  volume.  The  cubic  inch  is  practically  always 
used;  and  in  this  connection  it  is  written  thus,  inches3.  See  Table 
A,  page  52,  for  values  of  the  section  moduli  of  a  few  simple  sections. 

63.  Applications  of  the   First  Beam  Formula.     There  are 
three  principal  applications  of  equation  6,  which  will  now  be  ex- 
plained and  illustrated. 

64.  First  Application.     The  dimensions  of  a  beam  and  its 
manner  of  loading  and  support  are  given,  and  it  is  required  to 
compute  the  greatest  unit-tensile  and  compressive  stresses  in  the 
beam. 

This  problem  can  be  solved  by  means  of  equation  6,  written 
in  this  form, 

Q        MC  M 

or~ 


Unless  otherwise  stated,  we  assume  that  the  beams  are  uniform 
in  cross-section,  as  they  usually  are;  then  the  section  modulus 
(I-s-c)  is  the  same  for  all  sections,  and  S  (the  unit-fibre  stress  on 


60  STRENGTH  OF  MATERIALS 

the  remotest  fibre)  varies  just  as  M  varies,  and  is  therefore  greatest 
where  M  is  a  maximum.*  Hence,  to  compute  the  value  of  the 
greatest  unit-fibre  stress  in  a  given  case,  substitute  the  values  of 
the  section  modulus  and  the  maximum  bending  moment  in  the 
preceding  equation,  and  reduce. 

If  the  neutral  axis  is  equally  distant  from  the  highest  and  low- 
est fibres,  then  the  greatest  tensile  and  compressive  unit-  stresses 
are  equal,  and  their  value  is  S.  If  the  neutral  axis  is  unequally 
distant  from  the  highest  and  lowest  fibres,  let  c  denote  its  distance 
from  the  nearer  of  the  two,  and  S'  the  unit-fibre  stress  there. 
Then,  since  the  unit  -stresses  in  a  cross  -section  are  proportional  to 
the  distances  from  the  neutral  axis, 


If  the  remotest  fibre  is  on  the  convex  side  of  the  beam,  S  is  tensile 
and  S'  compressive;  if  the  remotest  fibre  is  on  the  concave  side,  S 
is  compressive  and  S'  tensile. 

Examples.  1.  A  beam  10  feet  long  is  supported  at  its  ends, 
and  sustains  a  load  of  4,000  pounds  two  feet  from  the  left  end 
(Fig.  37,  &).  If  the  beam  is  4  X  12  inches  in  cross-section  (the 
long  side  vertical  as  usual),  compute  the  maximum  tensile  and 
compressive  unit-stresses. 

The  section  modulus  of  a  rectangle  whose  base  and  altitude 
are  5  and  a  respectively  (see  Table  A,  page  52),  is  -J&&2;  hence, 
for  the  beam  under  consideration,  the  modulus  is 

i-  X  4  X  122  =  96  inches3. 

To  compute  the  maximum  bending  moment,  we  have,  first,  to  find 
the  dangerous  section.  This  section  is  where  the  shear  changes 
sign  (see  article  45);  hence,  we  have  to  construct  the  shear  dia- 
gram, or  as  much  thereof  as  is  needed  to  find  where  the  change  of 
sign  occurs.  Therefore  we  need  the  values  of  the  reaction. 
Neglecting  the  weight  of  the  beam,  the  moment  equation  with 
origin  at  C  (Fig.  37,  a)  is 

R,  X  10  -  4,000  X  8  =  0,  or  Rt  =  3,200  pounds 

*  NOTE.    Because  S  is  greatest  in  the  section  where  M  is  maximum,  this 
section  is  usually  called  the  "  dangerous  section  "  of  the  beam. 


STRENGTH  OF  MATERIALS 


61 


Then,  constructing  the  shear  diagram,  we  see  (Fig.  37,  5)  that  the 

change  of  sign  of  the  shear  (also  the  dangerous  section)  is  at  the 

load.     The  value  of  the  bending  moment  there  is 

3,200  X    2  =    6,400  foot-pounds, 

6,400  X  12  =  76,800  inch-pounds. 


or 


Substituting  in  equation  6',  we  find  that 
76,800 


8= 


96 
4oooltas. 


=  800 


inch. 


U-a'- 


c(a) 


B' 


c'(b) 


V  / 


Fig.  37. 

20     It  is  desired  to  take  into  account  the  weight  of  the  beam 
in  the  preceding  example,  supposing  the  beam  to  be  wooden. 
The  volume  of  the  beam  is 

—  X  10  =  3 J  cubic  feet ; 

and  supposing  the  timber  to  weigh  45  pounds  per  cubic  foot,  the 
beam  weighs  150  pounds  (insignificant  compared  to  the  load). 
The  left  reaction,  therefore,  is 

3,200 +(-i-X  150)  =  3,275; 


62  STRENGTH  OF  MATERIALS 

and  the  shear  diagram  looks  like  Fig.  37,  c,  the  shear  changing 
sign  at  the  load  as  before.  The  weight  of  the  beam  to  the  left  of 
the  dangerous  section  is  80  pounds;  hence  the  maximum  bending 
moment  equals 

3,275  X    2  -  30  X  1  =    6,520  foot-pounds, 
or  6,520  X  12  =  78,240  inch-pounds. 

Substituting  in  equation  6',  we  find  that 

78,240 
S  =  — q^ —  —  815  pounds  per  square  inch. 

The  weight  of  the  beam  therefore  increases  the  unit -stress  pro- 
duced by  the  load  at  the  dangerous  section  by  15  pounds  per 
square  inch. 

3.     A  T-bar  (see  Fig.  38)  8  feet  long  and  supported  at  each 
I  j  =^j      end,    bears    a  uniform    load    of    1,200 

pounds.     The  moment  of  inertia  of  its 
cross -section  with   respect  to   the  neu- 
"<0       tral  axis   being  2.42   inches*,  compute 
w       the  maximum  tensile  and  compressive 
-*-     unit-stresses  in  the  beam 


N 


Evidently    the    dangerous    section 

is  in  the  middle,  and  the  value  of  the  maximum  bending  moment 
(see  Table  B,  page  53,  Part  I)  is  J  WZ,  W  and  I  denoting  the  load 
and  length  respectively.  Here 

—  Wl  =  -g-  X  1,200  X  8  =  1,200  foot-pounds, 

or  1,200  X  12  =  14,400  inch-pounds. 

The  section  modulus  equals  2.42  -r-  2.28  =  1.06;  hence 

S  =      '   „    =  13,585  pounds  per  square  inch. 

This  is  the  unit-fibre  stress  on  the  lowest  fibre  at  the  middle  sec- 
tion,  and  hence  is  tensile.  On  the  highest  fibre  at  the  middle 
section  the  unit-stress  is  compressive,  and  equals  (see  page  60): 

S'  =  —  S  =  ^-ott  X  13,585  =  4,290  pounds  per  square  inch. 


STRENGTH  OF  MATERIALS  63 

EXAMPLES  FOR  PRACTICE. 

1.  A  beam  12  feet  long  and  6  X  12  inches  in  cross-section 
rests  on  end  supports,  and  sustains  a  load  of  3,000  pounds  in  the 
middle.      Compute    the   greatest   tensile   and   compressive  unit- 
stresses  in  the  beam,  neglecting  the  weight  of  the  beam. 

Ans.     750  pounds  per  square  inch. 

2.  Solve   the  preceding  example  taking  into   account  the 
weight  of  the  beam,  300  pounds 

Ans.     787.5  pounds  per  square  inch. 

3.  Suppose  that  a  built-in  cantilever  projects  5  feet  from  the 
wall  and  sustains  an  end  load  of  250  pounds.    The  cross-section  of 
the  cantilever  being  represented  in  Fig.  38,  compute  the  greatest 
tensile  and  compressive  unit-stresses,  and  tell  at  what  places  they 
occur.     (Neglect  the  weight.) 

j  Tensile,  4,471  pounds  per  square  inch. 

(  Compressive,  14,150       "         «        "         " 

4.  Compute  the  greatest  tensile  and  compressive  unit-stresses 
in  the  beam  of  Fig.  18,  a,  due  to  the  loads  and  the  weight  of  beam 
(400  pounds).     (A  moment  diagram  is  represented  in  Fig.  18,  5; 
for  description  see  example  2,  Art.  44,  p.  39.)      The  section  of 
the  beam  is  a  rectangle  8  X  12  inches. 

Ans.     580  pounds  per  square  inch. 

5.  Compute  the  greatest  tensile  and  compressive  unit-stresses 
in  the  cantilever  beam  of  Fig.  19,  #,  it  being  a  steel  I-beam  whose 
section  modulus  is  20.4  inches3.    (A  bending  moment  diagram  for 
it  is  represented  in  Fig.  19,  I;  for  description,  see  Ex.  3,  Art.  44.) 

Ans.     11,470  pounds  per  square  inch. 

6.  Compute  the  greatest  tensile  and  compressive  unit-stresses 
in  the  beam  of  Fig.  10,  neglecting  its  weight,  the  cross-sections 
being  rectangular  6  X  12  inches.      (See  example  for  practice  1, 
Art.  43.) 

Ans.     600  pounds  per  square  inch. 

65.  Second  Application.  The  dimensions  and  the  work- 
ing strengths  of  a  beam  are  given,  and  it  is  required  to  determine 
its  safe  load  (the  manner  of  application  being  given). 

This  problem  can  be  solved  by  means  of  equation  6  written 
in  this  form, 

M=  (6") 


64  STRENGTH  OF  MATERIALS 

"We  substitute  for  S  the  given  working  strength  for  the  ma- 
terial  of  the  beam,  and  for  I  and  c  their  values  as  computed  from 
the  given  dimensions  of  the  cross -section;  then  reduce,  thus 
obtaining  the  value  of  the  safe  resisting  moment  of  the  beam, 
which  equals  the  greatest  safe  bending  moment  that  the  beam  can 
stand.  "We  next  compute  the  value  of  the  maximum  bending 
moment  in  terms  of  the  unknown  load;  equate  this  to  the  value 
of  the  resisting  moment  previously  found;  and  solve  for  the 
unknown  load. 

In  cast  iron,  the  tensile  and  compressive  strengths  are  very 
different;  and  the  smaller  (the  tensile)  should  always  be  used  if 
the  neutral  surface  of  the  beam  is  midway  between  the  top  and 
bottom  of  the  beam;  but  if  it  is  unequally  distant  from  the  top 
and  bottom,  proceed  as  in  example  4,  following. 

Examples.  1.  A  wooden  beam  12  feet  long  and  6  X  12 
inches  in  cross -section  rests  on  end  supports.  If  its  working 
strength  is  800  pounds  per  square  inch,  how  large  a  load  uniformly 
distributed  can  it  sustain  ? 

The  section  modulus  is  \ba?,  b  and  a  denoting  the  base  and 
altitude  of  the  section  (see  Table  A,  page  52);  and  here 

i  la9  =r  i  x  6  X  122  =  144  inches3. 

0  0 

Hence  S  —  =  800  X  144  =  115,200  inch-pounds. 

c 

For  a  beam  on  end  supports  and  sustaining  a  uniform  load,  the 
maximum  bending  moment  equals  JW7  (see  Table  B,  page  55), 
"W  denoting  the  sum  of  the  load  and  weight  of  beam,  and  I  the 
length.  If  W  is  expressed  in  pounds,  then 

g-  Wl  =  g  W  X  12  foot-pounds  =  -^  W  X  144  inch-pounds. 

Hence,  equating  tl>Q  two  values  of.  maximum  bending  moment 
and  the  safe  resisting  moment,  we  get 

lw  X  144  —  115,200; 

o 

W-.11B  8-  6,400  pounds. 


STRENGTH  OF  MATERIALS  65 

The  safe  load  for  the  beam  is  6,400  pounds  minus  the  weight  of 
the  beam. 

2.  A  steel  I-beam  whose  section  modulus  is  20.4  inches8 
rests  on  end  supports  15  feet  apart.  Neglecting  the  weight  of  the 
beam,  how  large  a  load  may  be  placed  upon  it  5  feet  from  one  end, 
if  the  working  strength  is  16,000  pounds  per  square  inch? 

The  safe  resisting  moment  is 

RT 

-  =  16,000  X  20.4  =  326,400  inch-pounds; 
c 

hence  the  bending  moment  must  not  exceed  that  value.  The 
dangerous  section  is  under  the  load ;  and  if  P  denotes  the  unknown 
value  of  the  load  in  pounds,  the  maximum  moment  (see  Table  B, 
page  53,  Part  I)  equals  f  P  X  §  foot-pounds,  or  f  P  X  60  inch- 
pounds.  Equating  values  of  bending  and  resisting  moments, 
we  get 

|  P  X  60  =  326,400; 
o 

326,400  X  3 
or,  P  =      2'x  6Q       =  8,160  pounds. 

,3.  In  the  preceding  example,  it  is  required  to  take  into 
account  the  weight  of  the  beam.  375  pounds. 


5'- 


f 


W=  3751105. 
Fig.  39. 


As  we  do  not  know  the  value  of  the  safe  load,  we  cannot  con- 
struct  the  shear  diagram  and  thus  determine  where  the  dangerous 
section  is.  But  in  cases  like  this,  where  the  distributed  load  (the 
weight)  is  small  compared  writh  the  concentrated  load,  the  dan- 
gerous section  is  practically  always  where  it  is  under  the  concen- 
trated load  alone;  in  this  case,  at  the  load.  The  reactions  due  to 
the  weight  equal  £  X  375  =  187.5;  and  the  reactions  due  to  the 
load  equal  A  P  and  §  P,  P  denoting  the  value  of  the  load.  The 

1.  O  O  "  O 

larger  reaction  Rt   (Fig.  39)  hence  equals  §  P  +  187.5.     Since 


66  STRENGTH  OF  MATERIALS 

the  weight  of  the  beam  per  foot  is  375  -j-  15  =  25  pounds,  the 
maximum  bending  moment  (at  the  load)  equals 

(  |  P  +  187.5)  5  -  (25  X  5)  2J  = 

^  P  +  937.5  -  312.5  =  ^  P  +  625. 

This  is  in  foot-pounds  if  P  is  in  pounds. 

The  safe  resisting  moment  is  the  same  as  in  the  preceding 
illustration,  326,400  inch-pounds;  hence 

(-y  P  +  625)  12  =  326,400. 
Solving  for  P,  we  have 


,      10  P  —  79,725; 
or,  P  =  7,972.5  pounds. 

It  remains  to  test  our  assumption  that  the  dangerous  section 
is  at  the  load.  This  can  be  done  by  computing  Ex  (with  P  = 
7,972.5),  constructing  the  shear  diagram,  and  noting  where  the 
shear  changes  sign.  It  will  be  found  that  the  shear  changes  sign 
at  the  load,  thus  verifying  the  assumption. 

4.  A  cast-iron  built-in  cantilever  beam  projects  8  feet  from 
the  wall.  Its  cross-section  is  represented  in  Fig.  40,  and  the 

moment  of  inertia  with  respect  to 
the  neutral  axis  is  50  inches4;  the 
working  strengths  in  tension  and 
compression  are  2,000  and  9,000 
pounds  per  square  inch  respect- 
ively.  Compute  the  safe  uniform 
load  which  the  beam  can  sustain, 
neglecting  the  weight  of  the  bearn. 

The  beam  being  convex  up,  the  upper  fibres  are  in  tension 
and  the  lower  in  compression.  The  resisting  moment  (SI  -r-  c), 
as  determined  by  the  compressive  strength,  is 


STRENGTH  OF  MATERIALS  67 


=  100,000  inch-pounds; 

and  the  resisting  moment,  as  determined  by  the  tensile  strength,  is 

2,000  X  50 

-  ~-~  -  =  40.000  inch-pounds. 

6.U 

Hence  the  safe  resisting  moment  is  the  lesser  of  these  two,  or 
40,000  inch-pounds.  The  dangerous  section  is  at  the  wall  (see 
Table  B,  page  53),  and  the  value  of  the  maximum  bending 
moment  is  \  WZ,  W  denoting  the  load  and  I  the  length.  If  W  is 
in  pounds,  then 

M  =  i  W  X  8  foot-pounds  =  £  W  X  96  inch-pounds. 
Equating  bending  and  resisting  moments,  we  have 

-i-W  X  96  =  40,000; 

w=  40,000  X  2  ^833poundSt 

EXAMPLES  FOR  PRACTICE. 

1.  An  8  X  8  -inch  timber  projects  8  feet  from  a  wall.    If  its 
working  strength  is  1,000  pounds  per  square  inch,  how  large  an 
end  load  can  it  safely  sustain  ? 

Ans.     890  pounds. 

2.  A  beam  12  feet  long  and  8  X  16  inches  in  cross  -section, 
on  end  supports,  sustains  two  loads  P,  each  3  feet  from  its  ends 
respectively.    The  working  strength  being  1,000  pounds  per  square 
inch,  compute  P  (see  Table  B,  page  53). 

Ans.     9,480  pounds. 

3.  An  I-beam  weighing  25  pounds  per  foot  rests  on  end 
supports  20  feet  apart.     Its  section  modulus  is  20.4  inches3,  and 
its  working  strength  16,000  pounds  per  square  inch.     Compute 
the  safe  uniform  load  which  it  can  sustain. 

Ans.      10,880  pounds- 

66.  Third  Application.  The  loads,  manner  of  support, 
and  working  strength  of  beam  are  given,  and  it  is  required  to  de- 
termine the  size  of  cross-section  necessary  to  sustain  the  load 
safely,  that  is,  to  "design  the  beam." 


68  STRENGTH  OF  MATERIALS 

To  solve  this  problem,  we  use  the  first  beam  formula  (equation 
6),  written  in  this  form, 

JL    *L.  (6'") 

o       '  S 

We  first  determine  the  maximum  bending  moment,  and  then  sub- 
stitute its  value  for  M,  and  the  working  strength  for  S.  Then  we 
have  the  value  of  the  section  modulus  (I  -r-  <?„)  of  the  required 
beam.  Many  cross-sections  can  be  designed,  all  having  a  given 
section  modulus.  Which  one  is  to  be  selected  as  most  suitable  will 
depend  on  the  circumstances  attending  the  use  of  the  beam  and 
on  considerations  of  economy. 

.Examples.  1.  A  timber  beam  is  to  be  used  for  sustaining 
a  uniform  load  of  1,500  pounds,  the  distance  between  the  supports 
being  20  feet.  If  the  working  strength  of  the  timber  is  1,000  pounds 
per  square  inch,  what  is  the  necessary  size  of  cross-section  ? 

The  dangerous  section  is  at  the  middle  of  the  beam;  and  the 
maximum  bending  moment  (see  Table  B,  page  53)  is 

-i-WJ  =  ~  x  1,500  X  20  =  3,750  foot-pounds, 

or  3,750  X  12  =  45,000  inch-pounds. 

I         45,000 

Hence  • —  =     .,  AAA  =  45  inches3. 

c  1,UUU 

Now  the  section  modulus  of  a  rectangle  is  \bo?  (see  Table  A, 
page  54,  Part  I);  therefore,  \la?  =  45,  or  be?  =  270. 

Any  wooden  beam  (safe  strength  1,000  pounds  per  square 
inch)  whose  breadth  times  its  depth  square  equals  or  exceeds  270, 
is  strong  enough  to  sustain  the  load  specified,  1,500  pounds. 

To  determine  a  size,  we  may  choose  any  value  for  5  or  a,  and 
solve  the  last  equation  for  the  unknown  dimension.  It  is  best, 
however,  to  select  a  value  of  the  breadth,  as  1,  2,  3,  or  4  inches, 
and  solve  for  a.  Thus,  if  we  try  b  =  1  inch,  we  have 

a2  =  270,  or  a  =  16.43  inches. 

This  would  mean  a  board  1  X  18  inches,  which,  if  used,  would 
have  to  be  supported  sidewise  so  as  to  prevent  it  from  tipping  or 
"  buckling."  Ordinarily,  this  would  not  be  a  good  size. 

Next  try  J  =  2  inches;  we  have 

2  x  a2  =  270;  or  a  —  1/270  -f-  2  =  11.62  inches. 
This  would  require  a  plank  2  X  12,  a  better  proportion  than  the 
first.     Trying  5  =  £  inches,  we  have 


STRENGTH  OF  MATERIALS  69 

3  x  a?  =  270;  or  a  =  1/270  -*-  3  =  9.49  inches. 
This  would  require  a  plank  3  X  10  inches;  and  a  choice  between 
a  2  X  12  and  a  3  X  10  plank  would  be  governed  by  circumstances 
in  the  case  of  an  actual  construction. 

It  will  be  noticed  that  we  have  neglected  the  weight  of  the 
beam.  Since  the  dimensions  of  wooden  beams  are  not  fractional, 
and  we  have  to  select  a  commercial  size  next  larger  than  the  one 
computed  (12  inches  instead  of  11.62  inches,  for  example),  the 
additional  depth  is  usually  sufficient  to  provide  strength  for  the 
weight  of  the  beam.  If  there  is  any  doubt  in  the  matter,  we  can 
settle  it  by  computing  the  maximum  bending  moment  including 
the  weight  of  the  beam,  and  then  computing  the  greatest  uni*-fibre 
stress  due  to  load  and  weight.  If  this  is  less  than  the  safe  strength, 
the  section  is  large  enough;  if  greater,  the  section  is  too  small. 

Thus,  let  us  determine  whether  the  2  X  12-inch  plank  is 
strong  enough  to  sustain  the  load  and  its  own  weight.  Tb«  plank 
will  weigh  about  120  pounds,  making  a  total  load  of 

1,500  +  120  =  1,620  pounds. 
Hence  the  maximum  bending  moment  is 

JlWZ  ==  4-1,620  X  20  X  12  =  48,600  inch-pounds. 

o  o 

Since       -i  =  i  l>a2  =  -^-X  2  X  122  =  48,  and  S  =  j^L, 

S  =  -     '         =  1,013  pounds  per  square  inch. 

Strictly,  therefore,  the  2  X  12-inch  plank  is  not  large  enough;  but 
as  the  greatest  unit-stress  in  it  would  be  only  13  pounds  per  square 
inch  too  large,  its  use  would  be  permissible. 

2.  What  size  of  steel  I-beam  is  needed  to  sustain  safely  the 
loading  of  Fig.  9  if  the  safe  strength  of  the  steel  is  16,000  pounds 
per  square  inch  ? 

The  maximum  bending  moment  due  to  the  loads  was  found 
in  example  1,  Art.  43,  to  be  8,800  foot-pounds,  or  8,800  X  12  = 
105,600  inch-pounds. 

I        105,600 
—       T          = 


That  is,  an   I-beam   is  needed  whose  section  modulus  is  a  little 
larger  than  6.6,  to  provide  strength  for  its  own  weight. 


70 


STRENGTH  OF  MATERIALS 


To  select  a  size,  we  need  a  descriptive  table  of  I-beams,  such 
as  is  published  in  handbooks  on  structural  steel. 

Below  is  an  abridged  copy  of  such  a  table.  (The  last  two  columns  con- 
tain information  for  use  later.)  The  figure  illustrates  a  cross-section  of  an 
I-beam,  and  shows  the  axes  referred  to  in  the  table. 

It  will  be  noticed  that  two  sizes  are  given  for  each  depth; 
these  are  the  lightest  and  heaviest  of  each  size  that  are  made,  but 
intermediate  sizes  can  be  secured.  In  column  5  we  find  7.  3  as  the 
next  larger  section  modulus  than  the  one  required  (6,6);  and  this 
corresponds  to  a  12^-pound  6-inch  I-beam,  which  is  probably  the 
proper  size.  To  ascertain  whether  the  excess  (7.3-6.6  =  0.70) 
in  the  section  modulus  is  sufficient  to  provide  for  the  weight  of  the 
beam,  we  might  proceed  as  in  example  1.  In  this  case,  however, 
the  excess  is  quite  large,  and  the  beam  selected  is  doubtless  safe. 

TABLE  C. 

Properties  ot  Standard  I-Beams 


Section  of  beam,  showing  axes  1-1  and  2-2. 


1 

2 

3 

4 

5 

6 

Depth  of 
Beam, 
in  inches. 

Weight 
per  foot, 
in  pounds. 

Area  of  cross- 
section,  in 
square  inches. 

Moment  of 
inertia, 
axis  1—1. 

Section 
modulus, 
axis  1—1. 

Moment  of 
inertia, 
axis  2—2. 

3 

5.50 

1.63 

2.5 

1.7 

0.46 

3 

7.50 

2.21 

2.9 

1.9 

.60 

4 

7.50 

2.21 

6.0 

3.0 

.77 

4 

10.50 

3.09 

7.1 

3.6 

1.01 

5 

9.75 

2.87 

12.1 

4.8 

1.23 

5 

14.75 

4.34 

15.1 

6.1 

1.70 

6 

12.25 

3.61 

21.8 

7.3 

1.85 

6 

17.25 

£.07 

26.2 

8.7 

2.36 

7 

15.00 

4.42 

36.2 

10.4 

2.67 

7 

20.00 

5.88 

42.2 

12.1 

3.24 

8 

18.00 

5.33 

56.9 

14.2 

3.78 

8 

25.25 

7.43 

68.0 

17.0 

4.71 

9 

21.00 

6.31 

84.9 

18.9 

5.16 

9 

35.00 

10.29 

111.8 

24.8 

7.31 

10 

25.00 

7.37 

122.1 

24.4 

6.89 

10 

40.00 

11.76 

158.7 

31.7 

9.50 

12 

31.50 

9.26 

215.8 

36.0 

9.50 

12 

40.00 

11.76 

245.9 

41.0 

10.95 

15 

42.00 

12.48 

441.8 

58.9 

14.62 

15 

60.00 

17.65 

538.6 

71.8 

18.17 

18 

55.00 

15.93 

795.6 

88.4 

21.19 

18 

70.00 

20.59 

921.2 

102.4 

24.62 

20 

65.00 

19.08 

1,169.5 

117.0 

27.86 

20 

75.00 

22.06 

1,268.8 

126.9 

30.25 

24 

80.00 

23.32 

2,087.2 

173.9 

42.86 

24 

100.00 

29.41 

2,379.6 

198.3 

48.55 

STRENGTH  OF  MATERIALS  71 

EXAMPLES  FOR  PRACTICE. 

1.  Determine  the  size  of  a  wooden  beam  which  can  safely 
sustain  a  middle  load  of  2,000  pounds,  if  the  beam  rests  on  end 
supports  16  feet  apart,  and  its  working  strength  is  1,000  pounds 
per  square  inch.    Assume  width  6  inches. 

Ans.     6  X  10  inches. 

2.  What  sized  steel  I-beam    is  needed  to  sustain   safely  a 
uniform  load  of  200,000  pounds,  if  it  rests  on  end  supports  10 
feet  apart,  and  its  working  strength  is  16,000  pounds  per  square 
inch? 

Ans.        95-pound  24-incli. 

3.  What  sized  steel  I-beam  is  needed  to  sustain  safely  the 
loading  of  Fig.  10,  if  its  working  strength  is  16,000  pounds  per 
square  inch  ? 

Ans.     14.75-pound  5 -inch. 

67.  Laws  of  Strength  of  Beams.  The  strength  of  a  beam  is 
measured  by  the  bending  moment  that  it  can  safely  withstand ;  or, 
since  bending  and  resisting  moments  are  equal,  by  its  safe  resist, 
ing  moment  (SI  -f-  c).  Hence  the  safe  strength  of  a  beam  varies 
(1)  directly  as  the  working  fibre  strength  of  its  material,  azid  (2) 
directly  as  the  section  modulus  of  its  cross-section.  For  beams 
rectangular  in  cross-section  (as  wooden  beams),  the  section  modu- 
lus is  \bc?,  b  and  a  denoting  the  breadth  and  altitude  of  the 
rectangle.  Hence  the  strength  of  such  beams  varies  also  directly 
as  the  breadth,  and  as  the  square  of  the  depth.  Thus,  doubling 
the  breadth  of  the  section  for  a  rectangular  beam  doubles  the 
strength,  but  doubling  the  depth  quadruples  the  strength. 

The  safe  load  that  a  beam  can  sustain  varies  directly  as  its 
resisting  moment,  and  depends  on  the  way  in  which  the  load  is 
distributed  and  how  the  beam  is  supported.  Thus,  in  the  first 
four  and  last  two  cases  of  the  table  on  page  55, 

M  =  PZ,        hence  P  =--       SI  -f-  Ic, 

M  =  £  Wl,     "  W  =     2SI  -r-  Ic, 

M  =  |  P/,      "  P  =    4SI  -*•  Ic, 

M  =  £  WZ,     «  W  =    SSI  -4-  Ic, 

M  =  I  PI,      «  P  =    881  -f-  Ic, 

M  =  ^  WZ,  «  W  =  12SI  -*-  fo. 


72 


STRENGTH  OF  MATERIALS 


Therefore  the  safe  load  in  all  cases  varies  inversely  with  the 
length;  and  for  the  different  cases  the  safe  loads  are  as  1,  2,  4,  8, 
8,  and  12  respectively. 

Example.  What  is  the  ratio  of  the  strengths  of  a  plank  2  X 
10  inches  when  placed  edgewise  and  when  placed  flatwise  on  its 
supports  ? 

When  placed  edgewise,  the  section  modulus  of  the  plank  is 
\  X  2  X  102=--  33 J,  and  when  placed  flatwise  it  is  \  X  10  X  22  = 
6J-;  hence  its  strengths  in  the  two  positions  are  as  33J  to  6-| 
respectively,  or  as  5  to  1. 

EXAMPLE  FOR  PRACTICE. 

What  is  the  ratio  of  the  safe  loads  for  two  beams  of  wood, 
one  being  10  feet  long,  3x12  inches  in  section,  and  having  its  load 
in  the  middle;  and  the  other  8  feet  long  and  2x8  inches  in  section, 
with  its  load  uniformly  distributed. 

Ans.    As  28.8  to  21. 3 

68.  Modulus  of  Rupture.  If  a  beam  is  loaded  to  destruction, 
and  the  value  of  the  bending  moment  for  the  rupture  stage  is 
computed  and  substituted  for  M  in  the  formula  SI  -r-  c  =  M,  then 
the  value  of  S  computed  from  the  equation  is  the  modulus  of 
rupture  for  the  material  of  the  beam.  Many  experiments  have 
been  performed  to  ascertain  the  moduli  of  rupture  for  different 
materials  and  for  different  grades  of  the  same  material.  The  fol- 
owing  are  fair  values,  all  in  pounds  per  square  inch : 

TABLE  D. 

Moduli  of  Rupture. 


Timber: 
Spruce  

4,000—  7,000,  aver 
3,500      7,000, 
5,500    10,500, 
10,000    16,000, 
8,000    14,000, 
4,000    12,000, 
7,500    18,500, 
9,000    15,000, 

age     5,000 
4,500 
8,000 
12,500 
10,000 
8,000 
13,000 
11,500 

Hemlock  . 

White  pine. 

Long-leaf  pine.  .  .  . 
Short-leaf  pine.  .  . 
Douglas  spruce.  .  . 
White  oak  . 

Red  oak 

Stone  : 
Sandstone. 

400—  1,200, 
400      1,000. 
800      1,400. 

Limestone. 

Granite  

Cast  iron: 

One  and  one-half  to  two  and 
one-quarter  times  its  ulti- 
mate tensile  strength. 

Hard  steel: 

Varies  from  100,000  to  150,000 

STRENGTH  OF  MATERIALS  73 

Wrought  iron  and  structural  steels  have  no  modulus  of  rup- 
ture, as  specimens  of  those  materials  will  "  bend  double,"  but  not 
break.  The  modulus  of  rupture  of  a  material  is  used  principally 
as  a  basis  for  determining  its  working  strength.  The  factor  of 
safety  of  a  loaded  beam  is  computed  by  dividing  the  'modulus 
of  rupture  of  its  material  by  the  greatest  unit --fibre  stress  in 
the  beam. 

69.  The  Resisting  5hear.     The  shearing  stress  on  a  cross- 
section  of  a  loaded  beam  is  not  a  uniform  stress;  that  is,  it  is  not 
uniformly  distributed  over  the  section.     In  fact  the  intensity  or 
unit-stress  is  actually  zero  on  the  highest  and  lowest  fibres  of  a 
cross-section,  and  is  greatest,  in  such  beams  as  are  used  in  prac- 
tice, on  fibres  at  the  neutral  axis.     In  the  following  article  we 
explain  how  to  find  the  maximum  value  in  two  cases — cases  which 
are  practically  important. 

70.  Second   Beam   Formula.      Let    Ss  denote    the    average 
value  of  the  unit-shearing  stress  on  a  cross-section  of  a  loaded 
beam,  and  A  the  area  of  the  cross-section.     Then  the  value  of  the 
whole  shearing  stress  on  the  section  is  : 

Resisting  shear  =  Ss  A. 

Since  the  resisting  shear  and  the  external  shear  at  any  section  of  a 
beam  are  equal  (see  Art.  59), 

SSA  =  V.  (7) 

This  is  called  the  "  second  beam  formula  "     It  is  used  to  investi- 
gate and  to  design  for  shear  in  beams. 

In  beams  uniform  in  cross-section,  A  is  constant,  and  Ss  is 
greatest  in  the  section  for  which  Y  is  greatest.  Hence  the  great- 
est unit-shearing  stress  in  a  loaded  beam  is  at  the  neutral  axis  of 
the  section  at  which  the  external  shear  is  a  maximum.  There  is 
a  formula  for  computing  this  maximum  value  in  any  case,  but  it 
is  not  simple,  and  we  give  a  simpler  method  for  computing  the 
value  in  the  two  practically  important  cases: 

1.  In    wooden    beams  (rectangular  or  square  in  cross-section),  the 
greatest  unit-shearing  stress  in  a  section  is  50  per  cent  larger  than  the  average 
value  S§. 

2.  In  I-beams,  and  in  others  with  a  thin  vertical  web,  the  greatest 
unit-shearing  stress  in  a  section  practically  equals  S8,  as  given  by  equation  7, 
if  the  area  of  the  web  is  substituted  for  A. 


74  STRENGTH  OF  MATERIALS 

Examples.  1.  What  is  the  greatest  value  of  the  unit- 
shearing  stress  in  a  wooden  beam  12  feet  long  and  6x12  inches  in 
cross-section  when  resting  on  end  supports  and  sustaining  a  uni- 
form load  of  6,400  pounds  ?  (This  is  the  safe  load  as  determined 
by  working  fibre  stress;  see  example  1,  Art.  65.) 

The  maximum  external  shear  equals  one-half  the  load  (see 
Table  B,  page  53),  and  comes  on  the  sections  near  the  supports. 

Since  A  =  6  X  12  =  72  square  inches; 

3,200 
s  ~~     72       =        pounds  per  square  inch, 

and  the  greatest  unit- shearing  stress  equals 

3  3 

~2~  ^s  =  ~2~  ^  =  ^  pounds  per  square  inch. 

Apparently  this  is  very  insignificant;  but  it  is  not  negligible,  as 
is  explained  in  the  next  article. 

2.  A  steel  I-beam  resting  on  end  supports  15  feet  apart 
sustains  a  load  of  8,000  pounds  5  feet  from  one  end.  The  weight 
of  the  beam  is  375  pounds,  and  the  area  of  its  web  section  is  3.2 
square  inches.  (This  is  the  beam  and  load  described  in  examples 
2  and  3,  Art.  65.)  What  is  the  greatest  unit-shearing  stress  ? 

The  maximum  external  shear  occurs  near  the  support  where 
the  reaction  is  the  greater,  and  its  value  equals  that  reaction. 
Calling  that  reaction  R,  and  taking  moments  about  the  other  end 
of  the  beam,  we  have 

R  x  15  -  375  x  7-7T  -  8,000  x  10  =  0; 

therefore          15  R  =  80,000  +  2,812.5  ===  82,812.5; 
or,  R  =  5,520.8  pounds. 

Hence          Sg  =   '      —  =  1,725  pounds  per  square  inch. 

EXAMPLES  FOR  PRACTICE. 

1.  A  wooden  beam  10  feet  long  and  2  X  10  inches  in  cross- 
section  sustains  a  middle  load  of  1,000  pounds.  Neglecting  the 
weight  of  the  beam,  compute  the  value  of  the  greatest  unit-shearing 

stress. 

Ans.     37.5  pounds  per  square  inch. 


STRENGTH  OF  MATERIALS  75 

2.  Solve   the   preceding   example    taking  into  account   the 
weight  of  the  beam,  60  pounds. 

Ans.     40  pounds  per  square  inch. v 

3.  A  wooden  beam  12  feet  long  and  4  X  12  inches  in  cross- 
section  sustains  a  load  of  3,000  pounds  4  feet  from  one  end. 
Neglecting  the  weight  of  the  beam,  compute  the  value  of  the 
greatest  shearing  unit-stress. 

Ans.     G2.5  pounds  per  square  inch. 

71.  Horizontal  Shear.  It  can  be  proved  that  there  is  a 
shearing  stress  on  every  horizontal  section  of  a  loaded  beam.  An 
experimental  explanation  will  have  to  suffice  here.  Imagine  a 
pile  of  six  boards  of  equal  length  supported  so  that  they  do  not 
bend.  If  the  intermediate  supports  are  removed,  they  will  bend 
and  their  ends  will  not  be  flush  but  somewhat  as  represented  in 
Fig.  41.  This  indicates  that  the  boards  slid  over  each  other  during 
the  bending,  and  hence  there  was  a  rubbing  and  a  frictional  re- 
sistance exerted  by  the  boards  upon  each  other.  Now,  when  a 
solid  beam  is  being  bent,  there  is  an  exactly  similar  tendency  for 
the  horizontal  layers  to  slide  over  each  other;  and,  instead  of  a 
frictional  resistance,  there  exists  shearing  stress  on  all  horizontal 
sections  of  the  beam. 

In  the  pile  of  boards  the  amount  of  slipping  is  different  at 
different  places  between  any  two  boards,  being  greatest  near  the 
,supports  and  zero  midway  between  them.  Also,  in  any  cross- 
section  the  slippage  is  least  between  the  upper  two  and  lower  two 
boards,  and  is  greatest  between  the  middle  two.  These  facts  indi- 
cate that  the  shearing  unit- stress  on  horizontal  sections  of  a  solid 
beam  is  greatest  in  the  neutral  surface  at  the  supports. 

It  can  be  proved  that  at  any  place  in  a  beam  the  shearing 
unit-stresses  on  a  horizontal  and  on  a  vertical  section  are  equal. 


Fig.  41.  Fig.  42. 

It  follows  that  the  horizontal  shearing  unit- stress  is  greatest  at  the 
neutral  axis  of  the  section  for  which  the  external  shear  (V)  is  a 
maximum.  Wood  being  very  weak  in  shear  along  the  grain, 
timber  beams  sometimes  fail  under  shear,  the  "rupture"  being 


76  STRENGTH  OF  MATERIALS 

two  horizontal  cracks  along  the  neutral  surface  somewhat  as  rep- 
resented in  Fig.  42.  It  is  therefore  necessary,  when  dealing  with 
timber  beams,  to  give  due  attention  to  their  strength  as  determined 
by  the  working  strength  of  the  material  in  shear  along  the  grain. 

Example.  A  wooden  beam  3  X  10  inches  in  cross-section 
rests  on  end  supports  and  sustains  a  uniform  load  of  4,000  pounds 
Compute  the  greatest  horizontal  unit-stress  in  the  beam. 

The  maximum  shear  equals  one-half  the  load  (see  Table  B, 
page  55),  or  2,000  pounds.  Hence,  by  equation  7,  since  A  = 
3  X  10  =  30  square  inches, 

2,000  2 

s  =     30     =6677-  pounds  per  square  inch. 

This  is  the  average  shearing  unit-stress  on  the  cross-sections  near 
the  supports;  and  the  greatest  value  equals 

3  2 

-  X  66-7-  =  100  pounds  per  square  inch. 

According  to  the  foregoing,  this  is  also  1?he  value  of  the 
greatest  horizontal  shearing  unit-stress.  (If  of  white  pine,  for 
example,  the  beam  would  not  be  regarded  as  safe,  since  the  ulti- 
mate shearing  strength  along  the  grain  of  selected  pine  is  only 
about  400  pounds  per  square  inch.) 

72.  Design  of  Timber  Beams.  In  any  case  we  may  pro- 
ceed as  follows: — (1)  Determine  the  dimensions  of  the  cross- 
section  of  the  beam  from  a  consideration  of  the  fibre  stresses  as' 
explained  in  Art.  66.  (2)  With  dimensions  thus  determined,  com- 
pute the  value  of  the  greatest  shearing  unit-stress  from  the  formula, 

Greatest  shearing  unit-stress  —  -^-  Y  -=-  ab, 

where  Y  denotes  the  maximum  external  shear  in  the  beam,  and 
b  and  a  the  breadth  and  depth  of  the  cross-section. 

If  the  value  of  the  greatest  shearing  unit-stress  so  computed 
does  not  exceed  the  working  strength  in  shear  along  the  grain, 
then  the  dimensions  are  large  enough;  but  if  it  exceeds  that  value, 
then  a  or  &,  or  both,  should  be  increased  until  J.  V  -s-  ab  is  less 
than  the  working  strength.  Because  timber  beams  are  very  often 
"season  checked"  (cracked)  along  the  neutral  surface,  it  is  ad  vis- 


STRENGTH  OF  MATERIALS  77 

able  to  take  the  working  strength  of  wooden  beams,  in  shear  along 
the  grain,  quite  low.  One-twentieth  of  the  working  fibre  strength 
has  been  recommended*  for  all  pine  beams. 

If  the  working  strength  in  shear  is  taken  equal  to  one- 
twentieth  the  working  fibre  strength,  then  it  can  be  shown  that, 

1.  For  a  beam  on  end  supports  loaded  in  the  middle,  the  safe  load  de- 
pends on  the  shearing  or    fibre  strength  according  as  the  ratio  of  length  to 
depth  (I  -s-  a)  is  less  or  greater  than  10. 

2.  For  a  beam  on  end  supports  uniformly  loaded,  the  safe  load  depends 
on  the  shearing  or  fibre  strength  according  as  I  +  a  is  less  or  greater  than  20. 

Examples.  1.  It  is  required  to  design  a  timber  beam  to  sus- 
tain loads  as  represented  in  Fig.  11,  the  working  fibre  strength 
being  550  pounds  and  the  working  shearing  strength  50  pounds 
per  square  inch. 

The  maximum  bending  moment  (see  example  for  practice  3, 
Art.  43;  and  example  for  practice  2,  Art.  44)  equals  practically 
7,000  foot-pounds  or,  7,000  X  12  =  84,000  inch-pounds. 
Hence,  according  to  equation  6'", 


I       84,000 

=  -sra-  =  152.7  inches8. 
c          550 


Since  for  a  rectangle 


-     ba*  =  152.7,  or  ba*  =  916.2. 

Now,  if  we  let     5  =  4,  then  a2  =  229; 

or,  a  =  15.1  (practically  16)  inches. 

If,  again,  we  let  b  —  6,  then  a2  =  152.7; 

or  a  =  12.4  (practically  14)  inches. 

Either  of  these  sizes  wrill  answer  so  far  as  fibre  stress  is  concerned, 
but  there  is  more  "  timber  "  in  the  second. 

The  maximum  external  shear  in  the  beam  equals  1,556 
pounds,  neglecting  the  weight  of  the  beam  (see  example  3,  Art. 
87;  and  example  2,  Art.  38).  Therefore,  for  a  4  X  16-inch  beam, 

*  See  "Materials  of  Construction."  —  JOHNSON.    Page  55. 


78  STRENGTH  OF  MATERIALS 

3   ;  .  1,556 
Greatest  shearing  unit-  stress  =-gr  x  4  v  ifi 

=  36.5  pounds  per  square  inch; 
and  for  a  6  X  14-inch  beam,  it  equals 

3          1,556 
~9~  X  ^  -  TJ  =  27.7  pounds  per  square  inch. 

Since  these  values  are  less  than  the  working  strength  in  shear, 
either  size  of  beam  is  safe  as  regards  shear. 

If  it  is  desired  to  allow  for  weight  of  beam,  one  of  the  sizes 
should  be  selected.  First,  its  weight  should  be  computed,  then 
the  new  reactions,  and  then  the  unit-fibre  stress  may  be  com- 
puted as  in  Art.  64,  and  the  greatest  shearing  unit-stress  as  in  the 
foregoing.  If  these  values  are  within  the  working  values,  then 
the  size  is  large  enough  to  sustain  safely  the  load  and  the  weight 
of  the  beam. 

2.  What  is  the  safe  load  for  a  white  pine  beam  9  feet  long 
and  2x12  inches  in  cross-section,  if  the  beam  rests  on  end  supports 
and  the  load  is  at  the  middle  of  the  beam,  the  working  fibre 
strength  being  1,000  pounds  and  the  shearing  strength  50  pounds 
per  square  inch. 

The  ratio  of  the  length  to  the  depth  is  less  than  10;  hence 
the  safe  load  depends  on  the  shearing  strength  of  the  material 
Calling  the  load  P,  the  maximum  external  shear  (see  Table  B, 
page  53)  equals  -J-  P,  and  the  formula  for  greatest  shearing  unit 
stress  becomes 

3  -i-      P 

50  =  --x    2;  or  P  =  lj600  Pounds- 


EXAMPLES  FOR  PRACTICE. 

1.  "What  size  of  wooden  beam  can  safely  sustain  loads  as  in 
Fig.  12T  with  shearing  and  fibre  working  strength  equal  to  50  and 
1,000  pounds  per  square  inch  respectively  ? 

f:..     Ans.     6  X  12  inches 

2.  What  is  the  safe  load  for  a  wooden  beam  4  X  14  inches, 
and  18  feet  long,  if  the  beam  rests  on  end  supports  and  the  load 
is  uniformly  distributed,  with  working  strengths  as  in  example  1  ? 

Ans.     3,730  pounds 


STRENGTH  OF  MATERIALS  79 

73.  Kinds  of  Loads  and  Beams.     We  shall  now  discuss  the 
strength  of  beams  under  longitudinal  forces   (acting  parallel   to 
the    beam)    and  transverse  loads.     The   longitudinal   forces   are 
supposed  to  be  applied  at  the  ends  of  the  beams  and  along  the  axis* 
of    the  beam  in   each  case.     We  consider  only  beams  resting  on 
end  supports. 

The  transverse  forces  produce  bending  or  flexure,  and  the 
longitudinal  or  end  forces,  if  pulls,  produce  tension  in  the  beam; 
if  pushes,  they  produce  compression.  Hence  the  cases  to  be  con- 
sidered may  be  called  "  Combined  Flexure  and  Tension  "  and 
"  Combined  Flexure  and  Compression." 

74.  Flexure  and  Tension.     Let  Fig.  43,  <z,  represent  a  beam 
subjected  to  the  transverse  loads  L15  L2  and  L3,  and  to  two  equal 
end  pulls  P  and  P.    The  reactions  Rx  and  R2  are  due  to  the  trans- 
verse loads  and  can  be  computed  by  the  methods  of  moments  just 
as  though  there  were  no  end  pulls.     To  find  the  stresses  at  any 
cross  -section,    we  determine  those  due  to  the    transverse  forces 
(Lj,  L2,  L3,  R1  and  R2)  and  those  due  to  the  longitudinal;  then 
combine  these   stresses   to  get  the  total  effect  of  all  the  applied 
forces. 

The  stress  due  to  the  transverse  forces  consists  of  a  shearing 
stress  and  a  fibre  stress;  it  will  be  called  the  flexural  stress.  The 
fibre  stress  is  compressive  above  and  tensile  below.  Let  M  denote 
the  value  of  the  bending  moment  at  the  section  considered;  cl  and 
c2  the  distances  from  the  neutral  axis  to  the  highest  and  the  low- 
est fibre  in  the  section  ;  and  Sj  and  S2  the  corresponding  unit-fibre 
stresses  due  to  the  transverse  loads.  Then 


,  2 

bx  ==  —  y—  ;  and  !32  =  -y-  . 

The  stress  due  to  the  end  pulls  is  a  simple  tension,  and  it  equals 
P;  this  is  sometimes  called  the  direct  stress.  Let  S0  denote  the 
unit-tension  due  to  P,  and  A  the  area  of  the  cross-section;  then 

a       P 
b°  "  A- 

Both  systems  of  loads  to  the  left  of  a  section  between  Lj  and 


*  NOTE.    By  "  axis  of  a  beam  "  is  meant  the  line  through  the  centers  of 
gravity  of  all  the  cross-sections. 


80  STRENGTH  OF  MATERIALS 

L2  are  represented  in  Fig.  43,  &;  also  the  stresses  caused  by  them 
at  that  section.  Clearly  the  effect  of  the  end  pulls  is  to  increase  the 
tensile  stress  (on  the  lower 
fibres)  and  to  decrease  the 
compressive  stress  (on  the 
upper  fibres)  due  to  the  flex- 
u^e.  Let  Sc  denote  the  total 
(resultant)  unit- stress  on  the 
>per  fibre,  and  St  that  on 
e  lower  fibre,  due  to  all 
'\  the  forces  acting  on  the  beam. 
In  combining  the  stresses 
there  are  two  cases  to  con- 
sider: 

(1)  The  flexural  compressive  unit- stress  on  the  upper  fibre  is 
greater  than  the  direct  unit-stress;  that  is,  Sj  is  greater  than  S0. 
The  resultant  stress  on  the  upper  fibre  is 

Sc  =  Sj  -  S0  (compressive) ; 
and  that  on  the  lower  fibre  is 

St  =  S2  +  S0  (tensile). 

The  combined  stress  is  as  represented  in  Fig.  43,  c,  part  tensile 
and  part  compressive. 

(2)  The  flexural   compressive  unit-stress  is  less  than  the 
direct  unit-stress;  that  is,  Sj  is  less  than  S0.     Then  the  combined 
unit-stress  on  the  upper  fibre  is 

Sc  =  80-8!     (tensile); 
and  that  on  the  lower  fibre  is 

St  =  S2  +  S0     (tensile). 

The  combined  stress  is  represented  by  Fig.  43,  d,  and  is  all 
tensile. 

Example.  A  steel  bar  2x6  inches,  and  12  feet  long,  is  sub- 
jected to  end  pulls  of  45,000  pounds.  It  is  supported  at  each 
end,  and  sustains,  as  a  beam,  a  uniform  load  of  6,000  pounds. 
It  is  required  to  compute  the  combined  unit-fibre  stresses. 

Evidently  the  dangerous  section  is  at  the  middle,  and  M  = 

• ;  that  is, 


STRENGTH  OF  MATERIALS  81 

M  =  —  X  6,000  X  12  =  9,000  foot-pounds, 

8 

or  9,000  X  12  =  108,000  inch-pounds. 

The  bar  being  placed  with  the  six-inch  side  vertical, 
c1  =  c2  =  3  inches,  and 
I=JLx2x63  =  36  inches4.     (See  Art.  52.) 

I/O 

108,000  X  3 

Hence       S,  =  S2  = —^ —  -  =  9,000  pounds  per  square  inch. 

ob 

Since  A  =  2  X  6  =  12  square  inches, 

45,000 

fe0  =  — ^— —    =  o,7oU  pounds  per  square  men. 
l<o 

The  greatest  value  of  the  combined  compressive  stress  is 

S,  -  S0  =  9,000  -  3,750  =  5,250  pounds  per  square  inch, 
and  it  occurs  on  the  upper  fibres  of  the  middle  section.    The  great- 
est value  of  the  combined  tensile  stress  is 

S2  +  S0  =  9,000  +  3,750  =  12,750  pounds  per  square  inch, 
and  it  occurs  on  the  lowest  fibres  of  the  middle  section. 

EXAHPLE    FOR    PRACTICE. 

Change  the  load  in  the  preceding  illustration  to  one  of  6,000 
pounds  placed  in  the  middle,  and  then  solve. 

A        (  Sc  =  14,250  pounds  per  square  inch. 
QS'  j  St  =  21,750       «         «       «        « 

75.  Flexure  and  Compression.  Imagine  the  arrowheads  on 
P  reversed;  then  Fig.  43,  «,  will  represent  a  beam  under  com- 
bined flexural  and  compressive  stresses.  The  flexural  unit- stresses 
are  computed  as  in  the  preceding  article.  The  direct  stress  is  a 
compression  equal  to  P,  and  the  unit-stress  due  to  P  is  computed 
as  in  the  preceding  article.  Evidently  the  effect  of  P  is  to  increase 
the  compressive  stress  and  decrease  the  tensile  stress  due  to  the 
flexure.  In  combining,  we  have  two  cases  as  before: 

(1)  The  flexural  tensile  unit-stress  is  greater  than  the 
direct  unit-stress;  that  is,  S2  is  greater  than  S0.  Then  the  com- 
bined  unit-stress  on  the  lower  fibre  is 


82  STRENGTH  OF  MATERIALS 

St  =  S2  -  S0  (tensile) ; 
and  that  on  the  upper  fibre  is 

Sc  =  S,  -f  S0  (compressive). 

The  combined  fibre  stress  is  represented  by  Fig.  44,  a,  and  is  part 
tensile  and  part  compressive. 

(2)  The  flexural  unit-stress  on  the  lower  fibre  is  less  than 
the  direct  unit-stress;  that  is,  S2  is  less  than  S0.  Then  the  com- 
bined  unit-stress  on  the  lower  fibre  is 

St  =  S0  -  S2  (compressive); 
and  that  on  the  upper  fibre  is 

Sc  =  S0  +  Sj  (compressive). 
The  combined  fibre  stress  is  represented  by 
Fig.  44,  £,  and  is  all  compressive. 

Example.  A  piece  of  timber  6x6 
inches,  and  10  feet  long,  is  subjected  to  end 
pushes  of  9,000  pounds.  It  is  supported  in 
a  horizontal  position  at  its  ends,  and  sustains 
a  middle  load  of  400  pounds.  Compute  the 
combined  fibre  stresses. 

Evidently  the  dangerous  section  is  at  the 
middle,  and  M  =  l  P/;  that  is,  Fig.  44. 

M  =  j  X  400  X  10  =  1,000  foot-pounds, 

or  1,000  X  12  —  12,000  inch-pounds. 

Since      cl  =  c2  =  3  inches,  and 

1        3        1 

1  r=  "To  k>#  ==  ~TT\  X  o  X  6   =  108  inches4, 
l/o  ±/4 

12,000  X  3  1 

bx  —  fe2  _         ^Qg  =         ~3~  pounds  per  square  inch, 

Since      A  =  6  X  6  =  36  square  inches, 

9  000 
S0  =      '       =  250  pounds  per  square  inch. 


Hence  the  greatest  value  of  the  combined  compressive  stress  is 
So  +  Sa  =  333 -g-  -f  250  =  583-q-  pounds  per  square  inch. 


STRENGTH  OF  MATERIALS 


83 


Ans. 


It  occurs  on  the  upper  fibres  of  the  middle  section.     The  greatest 
value  of  the  combined  tensile  stress  is 

S2  -  S0  =  333-g-  -  250  =  83-^-  pounds  per  square  inch. 
It  occurs  on  the  lowest  fibres  of  the  middle  section. 

EXAMPLE  FOR  PRACTICE. 

Change  the  load  of  the  preceding  illustration  to  a  uniform 
load  and  solve. 

Sc  =  417  pounds  per  square  inch. 

St  ==    83       «         «        «         "      (compression). 

76.  Combined  Flexural  and  Direct  Stress  by  flore  Exact 
Formulas.  The  results  in  the  preceding  articles  are  only  approxi- 
mately correct.  Imagine  the 
beam  represented  in  Fig.  45,  a, 
to  be  first  loaded  with  the  trans- 
verse loads  alone.  They  cause 
the  beam  to  bend  more  or  less, 
and  produce  certain  flexural 
stresses  at  each  section  of  the 
beam.  Now,  if  end  pulls  are 
applied  they  tend  to  straighten 
the  beam  and  hence  diminish  the  flexural  stresses.  This  effect 
of  the  end  pulls  wTas  omitted  in  the  discussion  of  Art.  74,  and 
the  results  there  given  are  therefore  only  approximate,  the 
value  of  the  greatest  combined  fibre  unit-stress  (St)  being  too 
large.  On  the  other  hand,  if  the  end  forces  are  pushes,  they  in- 
crease the  bending,  and  therefore  increase  the  flexural  fibre  stresses 
already  caused  by  the  transverse  forces  (see  Fig.  45,  b).  The 
results  indicated  in  Art.  75  must  therefore  in  this  case  also  be 
regarded  as  only  approximate,  the  value  of  the  greatest  unit- 
fibre  stress  (Sc)  being  too  small. 

For  beams  loaded  in  the  middle  or  with  a  uniform  load,  the 
following  formulas,  which  take  into  account  the  flexural  effect  of 
the  end  forces,  may  be  used : 

M  denotes  bending  moment  at  the  middle  section  of  the  beam' 
I  denotes  the  moment  of  inertia  of  the  middle  section  with 
respect  to  the  neutral  axis; 


84  STRENGTH  OF  MATERIALS 

S1?  S2,  c1  and  c2  have  the  same  meanings  as  in  Arts.  74  and 
75,  but  refer  always  to  the  middle  section ; 

I  denotes  length  of  the  beam ; 

E  is  a  number  depending  on  the  stiffness  of  the  material,  the 
average  values  of  which  are,  for  timber,  1,500,000;  and  for  struc- 
tural steel  30,000,000.* 

S= 


~  10E 

The  plus  sign  is  to  be  used  when  the  end  forces  P  are  pulls,  and 
the  minus  sign  when  they  are  pushes. 

It  must  be  remembered  that  St  and  S2  are  flexural  unit- 
stresses.  The  combination  of  these  and  the  direct  unit-stress  is 
made  exactly  as  in  articles  74  and  75. 

Examples.  1.  It  is  required  to  apply  the  formulas  of  this 
article  to  the  example  of  article  74. 

As  explained  in  the  example  referred  to,  M  =  108,000  inch- 
pounds;  Cj=  c2=  3  inches;  and  I  =  36  inches4. 
Now,  since  I  =  12  feet  =  144  inches, 

108,000  X  3  324,000 

8'  =  S*  =  45,000  X  144*       =  36  +  Ml  ==  8'284   P°unda 

^  10  X  30,000,000 

per  square  inch,  as  compared  with  9,000  pounds  per  square  inch, 
the  result  reached  by  the  use  of  the  approximate  formula. 
As  before,  S0  —  3,750  pounds  per  square  inch;  hence 

Sc  =  8,284-  3,750  =  4,534  pounds  per  square  inch; 
and     St  =  8,284  +  3,750  =  12,034    "        "        "        « 

2.  It  is  required  to  apply  the  formulas  of  this  article  to  the 
example  of  article  75. 

As  explained  in  that  example, 

M  =  12,000  inch-pounds; 
c1  =  <?2  =.  3  inches,  and  I  =  108  inches4. 
Now,  since  I  =  120  inches, 

12,000  X  3  36,000 

•  362 


10  X  1,500,000 
*  NOTE.    This  quantity  "  E  "  is  more  fully  explained  in  Article  95. 


STRENGTH  OF  MATERIALS  85 

per  square  inch,  as  compared  with  333J  pounds  per  square  inch, 
the  result  reached  by  use  of  the  approximate  method. 
As  before,  S0  =  250  pounds  per  square  inch;  hence 

Sc  =  362  -f-  250  =  612  pounds  per  square  inch;  and 
St  ==  362  -  250  =±=  112        «         «        «         «     . 

EXAMPLES    FOR    PRACTICE. 

1.  Solve  the  example  for  practice  of  Art.  74  by  the  formulas 
of  this  article. 

A  n  *    $  Sc  =  12,820  pounds  per  square  inch. 
1S'  I  St  =  20,320       «         «        « 

2.  Solve  the  example  for  practice  of  Art.  75  by  the  formulas 
of  this  article. 

Ans     i  ^c  ~  ^^  pounds  per  square  inch. 

|  St  =    70       "         «        «         «      (compression). 

STRENGTH    OF    COLUHNS. 

A  stick  of  timber,  a  bar  of  iron,  etc.,  when  used  to  sustain 
end  loads  which  act  lengthwise  of  the  pieces,  are  called  columns, 
posts,  or  struts  if  they  are  so  long  that  they  would  bend  before 
breaking.  When  they  are  so  short  that  they  would  not  bend 
before  breaking,  they  are  called  short  blocks,  and  their  compres- 
sive  strengths  are  computed  by  means  of  equation  1.  The  strengths 
of  columns  cannot,  however,  be  so  simply  determined,  and  we  now 
proceed  to  explain  the  method  of  computing  them. 

77.  End  Conditions.  The  strength  of  a  column  depends  in 
part  on  the  way  in  which  its  ends  bear,  or  are  joined  to  other 
parts  of  a  structure,  that  is,  on  its  "  end  conditions."  There  are 
practically  but  three  kinds  of  end  conditions,  namely: 

1.  "  Hinge  "  or  "  pin  "  ends, 

2.  "Flat"  or  "square"  ends,  and 

3.  "Fixed"  ends. 

(1)  When  a  column  is  fastened  to  its  support  at  one  end  by 
means  of  a  pin  about  which  the  column  could  rotate  if  the  other 
end  were  free,  it  is  said  to  be  "  hinged "  or  "  pinned "  at  the 
former  end.    Bridge  posts  or  columns  are  often  hinged  at  the  ends. 

(2)  A  column  either  end  of  which  is  flat  and  perpendicular 
to  its  axis  and  bears  on  other  parts  of  the  structure  at  that  surface, 

square"  at  that  end. 


86 


STRENGTH  OF  MATERIALS 


(3)  Columns  are  sometimes  riveted  near  their  ends  directly 
to  other  parts  of  the  structure  and  do  not  bear  directly  on  their 
ends;  such  are  called  "  fixed  ended."  A  column  which  bears  on  its 
flat  ends  is  often  fastened  near  the  ends  to  other  parts  of  the  struc- 
ture, and  such  an  end  is  also  said  to  be  "  fixed."  The  fixing  of  an 
end  of  a  column  stiffens  and  therefore  strengthens  it  more  or  less, 
but  the  strength  of  a  column  with  fixed  ends  is  computed  as 
though  its  ends  were  flat.  Accordingly  we  have,  so  far  as  strength 
is  concerned,  the  following  classes  of  columns  : 

78.  Classes  of  Columns.  (1)  Both  ends  hinged  or  pinned; 
(2)  one  end  hinged  and  one  flat;  (3)  both  ends  flat. 

Other  things  being  the  same,  columns  of  these  three  classes 
are  unequal  in  strength.  Columns  of  the  first  class  are  the 
weakest,  and  those  of  the  third  class  are  the  strongest. 


_A_ 

=• 

_A_ 

B 

.A. 

1 

T" 

A 

i 

B 

a- 


Fig.  46. 


|B 
b 


79.  Cross=sections  of  Columns.     Wooden  columns  are  usu- 
ally solid,  square,  rectangular,  or  round  in  section ;  but  sometimes 
they  are  "  built  up  "  hollow.     Cast-iron  columns  are  practically 
always  made  hollow,  and  rectangular  or  round  in  section.     Steel 
columns  are  made  of  single  rolled  shapes — angles,  zees,  channels, 
etc.;  but  the  larger  ones  are  usually  "  built  up"  of  several  shapes. 
Fig.  46,  &,  for  example,  represents  a  cross -section  of  a  "  Z-bar" 
column;  and  Fig.  46,  &,  that  of  a  "  channel "  column. 

80.  Radius  of  Gyration.     There  is  a  quantity  appearing  in 
almost  all  formulas  for  the  strength  of  columns,  which  is  called 
u  radius  of  gyration."     It  depends  on  the  form  and  extent  of  the 
cross-section  of  the  column,  and  may  be  defined  as  follows: 


STRENGTH  OF  MATERIALS  87 

The  radius  of  gyration  of  any  plane  figure  (as  the  section  of  a  column) 
with  respect  to  any  line,  is  such  a  length  that  the  square  of  this  length  mul- 
tiplied by  the  area  of  the  figure  equals  the  moment  of  inertia  of  the  figure 
with  respect  to  the  given  line. 

Thus,  if  A  denotes  the  area  of  a  figure;  I,  its  moment  of  in- 
ertia  with  respect  to  some  line;  and  r,  the  radius  of  gyration 
with  respect  to  that  line;  then 

r*A.  =  I',orr  =  I/I  H-  A.  (9) 

In  the  column  formulas,  the  radius  of  gyration  always  refers  to  an 
axis  through  the  center  of  gravity  of  the  cross-section,  and  usually 
to  that  axis  with  respect  to  which  the  radius  of  gyration  (and  mo. 
ment  of  inertia)  is  least.  (For  an  exception,  see  example  3, 
Art.  83.)  Hence  the  radius  of  gyration  in  this  connection  is  often 
called  for  brevity  the  "  least  radius  of  gyration,"  or  simply  the 
"  least  radius." 

Examples.  1.  Show  that  the  value  of  the  radius  of  gyration 
given  for  the  square  in  Table  A,  page  52,  is  correct. 

The  moment  of  inertia  of  the  square  with  respect  to  the  axis 
is  TV^*»  Since  A  =  #2,  then,  by  formula  9  above, 

r  -  J17I7  =  JITa,  =  a  J  J. 

2.  Prove  that  the  value  of  the  radius  of  gyration  given  for 
the  hollow  square  in  Table  A,  page  54,  is  correct. 

The  value  of  the  moment  of  inertia  of  the  square  with  respect 
to  the  axis  is  T^  (a*  -  a').  Since  A  =  a2  -  a*, 


EXAHPLE  FOR  ^PRACTICE. 

Prove  that  the  values  of  the  radii  of  gyration  of  the  other  fig- 
ures given  in  Table  A,  page  52,  are  correct.  The  axis  in  each 
case  is  indicated  by  the  line  through  the  center  of  gravity. 

81.  Radius  of  Gyration  of  Built=up  Sections.  The  radius  of 
gyration  of  a  built-up  section  is  computed  similarly  to  that  of  any 
other  figure.  First,  we  have  to  compute  the  moment  of  inertia  of 


88  STRENGTH  OF  MATERIALS 

the  section,  as  explained  in  Art.  54;  and  then  we  use  formula  9,  as 
in  the  examples  of  the  preceding  article. 

Example.  It  is  required  to  compute .  the  radius  of  gyration 
of  the  section  represented  in  Fig.  30  (page  52)  with  respect  to  the 
axis  AA. 

In  example  1,  Art.  54,  it  is  shown  that  the  moment  of  inertia 
of  the  section  with  respect  to  the  axis  AA  is  429  inches4.  The 
area  of  the  whole  section  is 

2  X  6.03  +  7  =  19.06; 
hence  the  radius  of  gyration  r  is 

f429~ 
\  153)6  = 


4.74  inches. 


EXAMPLE  FOR  PRACTICE. 

Compute  the  radii  of  gyration  of  the  section  represented  in 
Fig.  31,  #,  with  respect  to  the  axes  AA  and  BB.  (See  examples 
for  practice  1  and  2,  Art.  54.) 

A  (  2.87  inches. 

Ans.     < 

\  2.09      « 

82.  Kinds  of  Column  Loads.     When  the  loads  applied  to  a 
column  are  such  that  their  resultant  acts  through  the  center  of 
gravity  of  the  top  section  and  along  the  axis  of  the  column,  the 
column  is  said  to  be  centrally  loaded.     When  the  resultant  of  the 
loads  does    not   act  through  the    center   of   gravity   of    the   top 
section,  the  column  is  said  to  be  eccentrically  loaded.     All  the 
following  formulas  refer  to  columns  centrally  loaded. 

83.  Rankine's  Column  Formula.     When  a  perfectly  straight 
column  is  centrally  loaded,  then,  if  the  column  does  not  bend  and 
if  it  is  homogeneous,  the  stress  on  every  cross-section  is  a  uniform 
compression.     If  P  denotes  the  load  and  A  the  area  of  the  cross- 
section,  the  value  of  the  unit-compression  is  P  -r-  A. 

On  account  of  lack  of  straightness  or  lack  of  uniformity  in 
material,  or  failure  to  secure  exact  central  application  of  the  load, 
the  load  P  has  what  is  known  as  an  "  arm  "  or  "  leverage  "  and 
bends  the  column  more  or  less.  There  is  therefore  in  such  a 
column  a  bending  or  flexural  stress  in  addition  to  the  direct  com- 
pressive  stress  above  mentioned ;  this  bending  stress  is  compressive 


STRENGTH  OF  MATERIALS  89 

on  the  concave  side  and  tensile  on  the  convex.  The  value  of  the 
stress  per  unit-area  (unit-stress)  on  the  fibre  at  the  concave  side, 
according  to  equation  6,  is  Me  -j-  I,  where  M  denotes  the  bending 
moment  at  the  section  (due  to  the  load  on  the  column),  c  the 
distance  from  the  neutral  axis  to  the  concave  side,  and  I  the 
moment  of  inertia  of  the  cross-section  with  respect  to  the  neutral 
axis.  (Notice  that  this  axis  is  perpendicular  to  the  plane  in 
which  the  column  bends.) 

The  upper  set  of  arrows  (Fig.  47)  represents  the  direct  com- 
pressive  stress;  and  the  second  set  the  bending  stress  if  the  load 
is  not  excessive,  so  that  the  stresses  are  within  the  elastic  limit  of 
the  material.  The  third  set  represents  the  combined  stress  that 
actually  exists  on  the  cross-section.  The  greatest  combined  unit- 
stress  evidently  occurs  on  the  fibre  at  the  concave  side  and  where 
the  deflection  of  the  column  is  greatest.  The 
stress  is  compressive,  and  its  value  S  per  unit- 
area  is  given  by  the  formula, 

P         Me 


=  TT    T- 

Now,   the   bending   moment  at   the  place  of 
greatest  deflection   equals    the  product  of  the 
load  P  and  its  arm  (that  is,  the  deflection). 
^*  Calling  the  deflection  d,  we  have  M  =  P</;  and 
this  value  of  M,  substituted  in  the  last  equa 
f^r^^^^^^—-   tion,  gives 


Fig.  47. 

Let  r  denote  the  radius  of  gyration  of  the  cross-section  with  respec 
to  the  neutral  axis.  Then  I  =  Ar2  (see  equation  9);  and  this 
value,  substituted  in  the  last  equation,  gives 

P        Rfo        P  do 

=  X  +  A?  -    ~K  "  +  ~?}- 

According  to  the  theory  of  the  stiffness  of  beams  on  end  sup- 
ports, deflections  vary  directly  as  the  square  of  the  length  Z,  and  in- 
versely as  the  distance  c  from  the  neutral  axis  to  the  remotest  fibre 
of  the  cross-section.  Assuming  that  the  deflections  of  columns 


90 


STRENGTH  OF  MATERIALS 


follow  the  same  laws,  we  may  write  d  =  ~k  (I2  -=-  c),  where  ~k  is 
some  constant  depending  on  the  material  of  the  column  and  on  the 
end  conditions.  Substituting  this  value  for  d  in  the  last  equation, 
we  find  that 


and 


P 

x  = 

P  = 


SA 


(10) 


Each  of  these  (usually  the  last)  is  known  as  "  Rankme's  formula." 

For  mild-steel  columns  a  certain  large  steel  company  uses  S  =  50,000 
pounds  per  square  inch,  and  the  following  values  of  Jc: 

1.  Columns  with  two  pin  ends,  k  =  1  -r-  18,000. 

2.  "  "    one  flat  and  one  pin  end,     k  =  1  -f-  24,000. 

3.  "  "     both  ends  flat,  fc  =  1  -*-  36,000. 

With  these  values  of  S  and  fc,  P  of  the  formula  means  the  ultimate  load, 
that  is,  the  load  causing  failure.  The  safe  load  equals  P  divided  by  the 
selected  factor  of  safety— a  factor  of  4  for  steady  loads,  and  5  for  moving 
loads,  being  recommended  by  the  company  referred  to.  The  same  unit  is  to 
be  used  for  Z  and  r. 

Cast-iron  columns  are  practically  always  made  hollow  with 
comparatively  thin  walls,  and  are  usually  circular  or  rectangular 
in  cross -section.  Tho  following  modifications  of  Rankine's  formula 
are  sometimes  used: 


For  circular  sections,  --r—  = 


For  rectangular  sections,  -^  = 


1,000  d2 


(ID") 


In  these  formulas  d  denotes  the  outside  diameter  of  the  circular  sec- 
tions or  the  length  of  the  lesser  side  of  the  rectangular  sections.  The  same 
unit  is  to  be  used  for  I  and  d. 

Examples.  1.  A  40-pound  10-inch  steel  I-beam  8  feet 
long  is  used  as  a  flat-ended  column.  Its  load  being  100,000 
pounds,  what  is  its  factor  of  safety  ? 

Obviously  the  column  tends  to  bend  in  a  plane  perpendicular 
to  its  web.  Hence  the  radius  of  gyration  to  be  used  is  the  one 


STRENGTH  OF  MATERIALS  91 

with  respect  to  that  central  axis  of  the  cross-section  which  is  in 
the  web,  that  is,  axis  2-2  (see  figure  accompanying  table,  page  72)  . 
The  moment  of  inertia  of  the  section  with  respect  to  that  axis, 
according  to  the  table,  is  9.50  inches4;  and  since  the  area  of  the 
section  is  11.76  square  inches, 

9'5°   --0808 
" 


"L76 

Now,  I  =  8  feet  =  96  inches;  and  since  k  =  1  -r-  36,000,  and  S  = 
50,000,  the  breaking  load  for  this  column,  according  to  Rankine's 
formula,  is 


p  =     50,000  X  11.76  _ 


446,790  pounds. 


1  + 


36,000  X  0.808. 


Since  the  factor  of  safety  equals  the  ratio  of  the  breaking  load  to 
the  actual  load  on  the  column,  the  factor  of  safety  in  this  case  is 

446,790 


2.  What  is  the  safe  load  for  a  cast-iron  column  10  feet  long 
with  square  ends  and  a  hollow  rectangular  section,  the  outside, 
dimensions  being  5x8  inches;  the  inner,  4x7  inches;  and  the 
factor  of  safety,  6  ? 

In  this  case  I  —  10  feet  =  120  inches;  A  =  5  X8-4  X  7 
=  12  square  -inches;  and  d  =  5  inches.  Hence,  according  to 
formula  10',  for  rectangular  sections,  the  breaking  load  is 

P  =   80.000X^18    =  610>000  pound, 

h  1,000  X  52 

Since  the  safe  load  equals  the  breaking  load  divided  by  the  factor 
of  safety,  in  this  case  the  safe  load  equals 

=  101,700  pounds. 

3.  A  channel  column  (see  Fig.  46,  $)  is  pin-ended,  the  pins 
being  perpendicular  to  the  webs  of  the  channel  (represented  by 
AA  in  the  figure),  and  its  length  is  16  feet  (distance  between  axes 


92  STRENGTH  OF  MATERIALS 

of  the  pins).  If  the  sectional  area  is  23.5  square  inches,  and  the 
moment  of  inertia  with  respect  to  AA  is  386  inches4  and  with 
respect  to  BB  214  inches4,  what  is  the  safe  load  with  a  factor  of 
safety  of  4  ? 

The  column  is  liable  to  bend  in  one  of  two  ways,  namely,  in 
the  plane  perpendicular  to  the  axes  of  the  two  pins,  or  in  the  plane 
containing  those  axes. 

(1)  For  bending  in  the  first  plane,  the  strength  of  the  col- 
umn is  to  be  computed  from  the  formula  for  a  pin  -ended  column. 
Hence,  for  this  case,  r2  =  386  -r-  23.5  =  16;  and  the  breaking 
load  is 

p= 


" 


18,000  X  16 
The  safe  load  for  this  case  equals  —  -  —  -^  -  =  260,400  pounds. 

A  i 

(2)  If  the  supports  of  the  pins  are  rigid,  then  the  pins 
stiffen  the  column  as  to  bending  in  the  plane  of  their  axes,  and  the 
strength  of  the  column  for  bending  in  that  plane  should  be  com- 
puted from  the  formula  for  the  strength  of  columns  with  flat  ends. 
Hence,  r*  =  214  -f-  23.5  =  9.11,  and  thebreaking  load  is 

P  =  --  '  nf^v  iov     =  1,056,000  pounds. 


36,000  X  9.11 
The  safe  load  for  this  case  equals  -  -  =  264,000  pounds. 

EXAMPLES   FOR    PRACTICE. 

1.  A  40-pound  12-inch  steel  I-beam  10  feet  long  is  used  as 
a  column  with  flat    ends  sustaining  a  load  of  100,000  pounds. 

What  is  its  factor  of  safety? 

Ans.     4.1 

2.  A  cast-iron    column    15    feet    long    sustains   a   load    of 
150,000  pounds.     Its  section  being  a  hollow  circle,  9  inches  out- 
side and  7  inches  inside  diameter,  what  is  the  factor  of  safety? 

Ans.     8.9 

3.  A  steel  Z-bar  column  (see  Fig.  46,  a)  is  24  feet  long  and 
has  square  ends;  the  least  radius  of  gyration  of  its  cross-section  is 


STRENGTH  OF  MATERIALS 


93 


3.1  inches;  and  the  area  of  the  cross-section  is  24.5  square  inches. 
What  is  the  safe  load  for  the  column  with  a  factor  of  safety  of  4  ? 

Ans.     247,000  pounds. 

4.  A  cast-iron  column  13  feet  long  has  a  hollow  circular 
cross-section   7  inches   outside    and    5J  inches    inside   diameter. 
What  is  its  safe  load  with  a  factor  of  safety  of  6  ? 

Ans.     121,142  pounds. 

5.  Compute   the    safe   load    for   a  40-pound  12-inch    steel 
I-beam  used  as  a  column  with  flat  ends,  its  length  being  17  feet. 
Use  a  factor  of  safety  of  5. 

Ans.     52,470  pounds. 

84.  Graphical  Representation  of  Column  Formulas.  Col- 
umn (and  most  other  engineering)  formulas  can  be  represented 
graphically.  To  represent  Rankine's  formula  for  flat-ended  mild- 
steel  columns, 

P  50,000 


36,000 

we  first  substitute  different  values  of  I  -r-  r  in  the  formula,  and 
solve  for  P  -r-  A.     Thus  we  find,  when 

Z  -  r  =    40,  P  -  A  =  47,900  ; 
I  +  r  =    80,  P  -  A  =  42,500  ; 
l  +  r  =  120,  P  -*-  A  =  35,750  ; 
etc.,  etc. 

Now,  if  these  values  of  I  -s-  r  be  laid  off  by  some  scale  on  a  line 
from  O,  Fig.  48,  and  the  corresponding  values  of  P  -r-  A  be  laid 


,1-r-r 


100 


zoo 


300 


Fig.  48. 

off  vertically  from  the  points  on  the  line,  we  get  a  series  of  points 
as  #,  &,  6',  etc.;  and  a  smooth  curve  through  the  points  a,  b,  ct 


94  STRENGTH  OF  MATERIALS 

etc.,  represents  the  formula.  Such  a  curve,  besides  representing 
the  formula  to  one's  eye,  can  be  used  for  finding  the  value  of 
P  -5-  A  for  any  value  of  I  -f-  r;  or  the  value  of  I  -f-  r  for  any  value 
of  P  -5-  A.  The  use  herein  made  is  in  explaining  other  column 
formulas  in  succeeding  articles. 

85.  Combination  Column  Formulas.  Many  columns  have 
been  tested  to  destruction  in  order  to  discover  in  a  practical  way 
the  laws  relating  to  the  strength  of  columns  of  different  kinds. 
The  results  of  such  tests  can  be  most  satisfactorily  represented 
graphically  by  plotting  a  point  in  a  diagram  for  each  test.  Thus, 
suppose  that  a  column  whose  I  -r-  r  was  80  failed  under  a  load  of 
276,000  pounds,  and  that  the  area  of  its  cross-section  was  7.12 
square  inches.  This  test  would  be  represented  by  laying  off  Oa, 
Fig.  49,  equal  to  80,  according  to  some  scale;  and  then  ab  equal  to 
276,000  -r-  7.12  (P  H-  A),  according  to  some  other  convenient 
scale.  The  point  b  would  then  represent  the  result  of  this  par- 
ticular test.  All  the  dots  in  the  figure  represent  the  way  in  which 
the  results  of  a  series  of  tests  appear  when  plotted. 

It  will  be  observed  at  once  that  the  dots  do  not  fall  upon  any 
one  curve,  as  the  curve  of  Rankine's  formula.     Straight  lines  and 


50000  •• 
AOOOO-  «- 
30OOO  •  • 
2OOOO-- 

10OOO 


,V*-r 


1OO  2OO  300 

Fig.  49. 

curves  simpler  than  the  curve  of  Rankine's  formula  have  been 
fitted  to  represent  the  average  positions  of  the  dots  as  determined 
by  actual  tests,  and  the  formulas  corresponding  to  such  lines  have 
been  deduced  as  column  formulas.  These  are  explained  in  the 
following  articles. 

86.    Straight-Line  and  Euler  Formulas.     It  occurred  to  Mr. 
T.  H.  Johnson  that  most  of  the  dots  corresponding  to  ordinary 


STRENGTH  OF  MATERIALS 


95 


lengths  of  columns  agree  with  a  straight  line  just  as  well  as  with 
a  curve.  He  therefore,  in  1886,  made  a  number  of  such  plats  or 
diagrams  as  Fig.  49,  fitted  straight  lines  to  them,  and  deduced  the 
formula  corresponding  to  each  line.  These  have  become  known 
as  "  straight-line  formulas,"  and  their  general  form  is  as  follows: 


— r-  =  S  -  m — , 

A  T 


(•I) 


P,  A,  Z,  and  r  having  meanings  as  in  Rankine's  formula  (Art.  83), 
and  S  and  m  being  constants  whose  values  according  to  Johnson 
are  given  in  Table  E  below. 

For  the  slender  columns,  another  formula  (Euler's,  long  since 
deduced)  was  used  by  Johnson.     Its  general  form  is — 

JP  n 

A 


(12) 


(I  +  rf 

n  being  a  constant  whose  values,  according  to  Johnson,  are  given 
in  the  following  table: 

TABLE  E. 

Data  for  Mild-Steel  Columns. 


8 

m 

Limit  (I  -i-  r) 

n 

Hinged  ends  

52,500 

220 

160 

444,000,000 

Flat  ends  

52,500 

180 

195 

666,000,000 

The  numbers  in  the  fourth  column  of  the  table  mark  the  point  of  divi- 
sion between  columns  of  ordinary  length  and  slender  columns.  For  the 
former  kind,  the  straight-line  formula  applies;  and  for  the  second,  Euler's. 
That  is,  if  the  ratio  /  •*•  r  for  a  steel  column  with  hinged  end,  for  example,  is 
less  than  160,  we  must  use  the  straight-line  formula  to  compute  its  safe  load, 
factor  of  safety,  etc.;  but  if  the  ratio  is  greater  than  160,  we  must  use  Euler's 
formula. 

For  cast-iron  columns  with  flat  ends,  S  =  34,000,  and  m  =  88;  and  since 
they  should  never  be  used  "  slender,"  there  is  no  use  of  Euler's  formula  for 
cast-iron  columns. 

The  line  AB,  Fig.  50,  represents  Johnson's  straight-line  for- 
mula; and  BC,  Euler's  formula.  It  will  be  noticed  that  the  two 
lines  are  tangent;  the  point  of  tangency  corresponds  to  the  "lim- 
iting value  "  I  -r-  r,  as  indicated  in  the  table. 

Examples.     1.     A  40-pound  10-in^.h  steel  I-beam  column  8 


96 


STRENGTH  OF  MATERIALS 


feet  long  sustains  a  load  of  100,000  pounds,  and  the  ends  are  flat. 
Compute  its  factor  of  safety  according  to  the  methods  of  this 
article. 

The  first  thing  to  do  is  to  compute  the  ratio  I  -r-  r  for  the 
column,  to  ascertain  whether  the  straight-line  formula  or  Euler's 

P-f-A 

50000 
4-OOOO- 
30OOO 
2OOOO- 
JOOOO 


100 


200 


300 


Fig.  50. 

formula  should  be  used.  From  Table  C,  on  page  70,  we  find  that 
the  moment  of  inertia  of  the  column  about  the  neutral  axis  of 
its  cross -section  is  9.50  inches4,  and  the  area  of  the  section  is 
11.76  square  inches.  Hence 


9.50 
1L76" 


=  0.81;  or  r  =  0.9  inch. 


Since  I  =  8  feet  =  96  inches, 

J_         96 
r 


0.9  -  106T 


This  value  of  I  -5-  r  is  less  than  the  limiting  value  (195)  indicated 
by  the  table  for  steel  columns  with  flat  ends  (Table  E,  p.  97),  and 
we  should  therefore  use  the  straight-line  formula;  hence 

=  52,500  -  180  X  106-jp 

or,          P  =  11.76  (52,500  -  180  X  106-?-)  =  391,600 pounds. 

This  is  the  breaking  load  for  the  column  according  to  the  straight, 
line  formula;  hence  the  factor  of  safety  is 

391,600 
100,000 


STRENGTH  OF  MATERIALS  97 

2.  Suppose  that  the  length  of  the  column  described  in  the 
preceding  example  were  16  feet.  What  would  its  factor  of  safety  be? 

Since  I  =  16  feet  =  192  inches;  and,  as  before,  r  =  0.9 
inch,  I  ~-  r  =  21  3  J.  This  value  is  greater  than  the  limiting 
value  (195)  indicated  by  Table  E  (p.  97)  for  flat-ended  steel  col- 
umns;  hence  Euler's  formula  is  to  be  used.  Thus 

P       _  666,000,000 
IL76  "        (213J)2     ' 

11.76  X  666,000,000 
or,  P  =  -  «      -  ^  172>100 


This  is  the  breaking  load;  hence  the  factor  of  safety  is 
172,100 


100,000 


1.7 


3.  What  is  the  safe  load  for  a  cast-iron  column  10  feet  long 
with  square  ends  and  hollow  rectangular  section,  the  outside 
dimensions  being  5x8  inches  and  the  inside  4x7  inches,  with  a 
factor  of  safety  of  6  ? 

Substituting  in  the  formula  for  the  radius  of  gyration  given 
in  Table  A,  page  52,  we  get 


r=     ,      8  X  53-7x  43 


12  (8  X  5  -  7  X  4) 
Since  I  =  10  feet  =  120  inches, 

1  12°         6122 

~       T96   =  6L22 

According  to  the  straight-line  formula  for  cast  iron,  A  being 
equal  to  12  square  inches, 


=  34?000  -  88  X  61.22; 

or,  P  =  12  (34,000  -  88  X  61.22)  =  343,360  pounds. 

This  being  the  breaking  load,  the  safe  load  is 

=  57,227  pounds. 


98  STRENGTH  OF  MATERIALS 

EXAMPLES  FOR  PRACTICE. 

1.  A  40-pound  12-inch  steel  I-beam  10  feet  long  is  used  as 
a  flat-ended  column.     Its  load  being  100,000  pounds,  compute 
the  factor  of  safety  by  the  formulas  of  this  article. 

Ans.     3.5 

2.  A  cast-iron  column   15    feet   long   sustains   a   load    of 
150,000  pounds.     Its  section  being  a  hollow  circle  of  9  inches 
outside  and  7  inches  inside  diameter,  compute  the  factor  of  safety 
by  the  straight-line  formula. 

Ans.     4.8 

3.  A  steel  Z-bar  column   (see  Fig.  46,  a]  is  24  feet  long 
and  has  square  ends;   the  least  radius  of  gyration  of  its  cross- 
section  is  3.1  inches;  and  the  area  of  the  cross-section  is   24.5 
square   inches.     Compute  the  safe  load  for  the  column  by  the 
formulas  of  this  article,  using  a  factor  of  safety  of  4. 

Ans.     219,000  pounds. 

4.  A  hollow  cast-iron  column  13  feet  long  has  a  circular 
cross-section,  and  is  7  inches  outside  and  5J  inches  inside   in 
diameter.     Compute  its  safe  load  by  the  formulas  of  this  article, 
using  a  factor  of  safety  of  6. 

Ans.     68,500  pounds 

5.  Compute  by  the  methods  of  this  article  the  safe  load  for 
a  40-pound  12  -inch  steel  I-beam  used  as  a  column  with  flat  ends,, 
If  the  length  is  17  feet  and  the  factor  of  safety  5. 

Ans.     35,100  pounds. 

87.  ParaboIa=EuIer  Formulas.  As  better  fitting  the  results 
of  tests  of  the  strength  of  columns  of  "  ordinary  lengths,"  Prof. 
J.  B.  Johnson  proposed  (1892)  to  use  parabolas  instead  of  straight 
lines.  The  general  form  of  the  "  parabola  formula  "  is 


P,  A,  I  and  r  having  the  same  meanings  as  in  Rankine's  formula, 
Art.  83;  and  S  and  m  denoting  constants  whose  values,  according 
to  Professor  Johnson,  are  given  in  Table  F  below. 

Like  the  straight-line  formula,  the  parabola  formula  should 
not  be  used  for  slender  columns,  but  the  following  (Euler's)  is 
applicable: 


STRENGTH  OF  MATERIALS 


99 


n 


the  values  of  n  (Johnson)  being  given  in  the  following  table 

TABLE  F. 

Data  for  Jlild  Steel  Columns. 


s 

m 

Limit  (I  H-  r) 

n 

Hinged  ends  

42,000 

0.97 

150 

456,000,000 

Flat  ends  .  . 

42.000 

0.62 

190 

712,000,000 

The  point  of  division  between  columns  of  ordinary  length  and  slender 
columns  is  given  in  the  fourth  column  of  the  table.  That  is,  if  the  ratio  Z-Hr 
for  a  column  with  hinged  ends,  for  example,  is  less  than  150,  the  parabola 
formula  should  be  used  to  compute  the  safe  load,  factor  of  safety,  etc.;  but 
if  the  ratio  is  greater  than  150,  then  Euler's  formula  should  be  used. 

The  line  AB,  Fig.  51,  represents  the  parabola  formula  ;  and  the  line 
BC,  Euler's  formula.  The  two  lines  are  tangent,  and  the  point  of  tangency 
corresponds  to  the  "  limiting  value"  l-^-r  of  the  table. 

For  wooden  columns  square  in  cross-section,  it  is  convenient  to  replace 
r  by  d,  the  latter  denoting  the  length  of  the  sides  of  the  square.  The  formula 
becomes 

'        : 


S  and  m  for  flat-ended  columns  of  various  kinds  of  wood  having  the  follow- 
ing  values  according  to  Professor  Johnson: 

For  White  pine,  8=2,500,  w  =  0.6; 

11    Short-leaf  yellow  pine,  8=3,300,  m  =  0.7j 

"    Long-leaf  yellow  pine,  8=4,000,  m  =  0.8; 

11    White   oak,  8=3,500,  w  =  0.8. 

The  preceding  formula  applies  to  any  wooden  column  whose  ratio,  l-r-d, 
is  less  than  60,  within  which  limit  columns  of  practice  are  included. 


10000- 


100  200 

Fig.  51. 


300 


Examples.     1.     A  40-pound  10-inch  steel   I-beam  column 


100  STRENGTH  OF  MATERIALS 

8  feet  long  sustains  a  load  of  100,000  pounds,  and  its  ends  are  flat. 
Compute  its  factor  of  safety  according  to  the  methods  of  this 
article. 

The  first  thing  to  do  is  to  compute  the  ratio  I  -f-  r  for  the 
column,  to  ascertain  whether  the  parabola  formula  or  Euler's  for- 
mula should  be  used.  As  shown  in  example  1  of  the  preceding 

article,  I  -s-  r  =  106§.     This  ratio  being  less  than  the  limiting 

-A         .j 
value,   190,  of  the  table,   we   shoula  use^he  ;parabola  formula. 

Hence,  since  the  area  of  the  cross  -"section-  is  11.76  square  inches 
(see  Table  C,  page  70),  .-.« 

-*-B  =  42,000  -0.62  (106f)«; 

or,  P  =  11.76  [42,000  -  0.62  (106f  )2]  =  410,970  pounds. 

This  is  the  breaking  load  according  to  the  parabola  formula;  hence 
the  factor  of  safety  is 

410,970  __  41 

100,000  " 

2.  A  white  pine  column  10  X  10  inches  in  cross-section  and 
18  feet  long  sustains  a  load  of  40,000  pounds.     What  is  its  factor 
of  safety  ? 

The  length  is  18  feet  or  216  inches;  hence  the  ratio  I  -f-  d  = 
21.6,  and  the  parabola  formula  is  to  be  applied. 
Now,  since  A  =  10  X  10  =  100  square  inches, 

j^-  =  2,500  -  0.6  X  21.6'; 

or,  P  =  100  (2,500  -  0.6  X  21.62)  =  222,000  pounds. 

This  being  the  breaking  load  according  to  the  parabola  formula, 
the  factor  of  safety  is 

222,000 

40,000   = 

3.  What  is  the  safe  load  for  a  long-leaf  yellow  pine  column 
12  X  12  inches   square  and  30  feet  long,  the  factor  of   safety 
being  5  ? 

The  length  being  30  feet  or  360  inches, 

1  -  -  36°      so- 

d  -  12"  ~  3°> 


STRENGTH  OF  MATERIALS , . , 101 


hence  the  parabola  formula  should  be  used.     Since  A  =  12  X  12 
=  144  square  inches, 

JL  =  4,000  -  0.8  X  302; 

or,  P  ==  144  (4,000  -  0.8  X  302)  =  472,320  pounds. 

This  being  the  breaking  load  according  to  the  parabola  formula, 
the  safe  load  is 

1J!  -  =  94,465  pounds. 

EXAMPLES   FOR   PRACTICE. 

1.  A  40-pound  12-inch  steel  I-beam  10  feet  long  is  used  as 
a  flat-ended  column.     Its  load  being  100,000  pounds,  compute  its 
factor  of  safety  by  the  formulas  of  this  article. 

Ans.     3.8 

2.  A  white  oak  column   15   feet   long  sustains  a  load  of 
30,000  pounds.      Its  section  being  8x8  inches,  compute   the 
factor  of  safety  by  the  parabola  formula. 

Ans.     6.6 

3.  A  steel  Z-bar  column  (see  Fig.  46,  a)  is  24  feet  long  and 
has  square  ends;  the  least  radius  of  gyration  of  its  cross-section 
is   3.1  inches;  and  the  area  of  its  cross-section  is   24.5    square 
inches.     Compute  the  safe  load  for  the  column  by  the  formulas 
of  this  article,  using  a  factor  of  safety  of  4. 

Ans.     224,500  pounds. 

4.  A  short-leaf  yellow  pine  column  14  X  14  inches  in  sec- 
tion is  20  feet  long.     What  load  can  it  sustain,  with  a  factor  of 
safety  of  6  ? 

Ans.   101,100  pounds. 

88.  "  Broken  Straight-Line  "  Formula.  A  large  steel  com- 
pany computes  the  strength  of  its  flat-ended  steel  columns  by  two 
formulas  represented  by  two  straight  lines  AB  and  BC,  Fig.  52. 
The  formulas  are 

X  =  48,000, 

and  -?-  =  68,400  -  228    £, 

P.  A,  I,  and  r  having  the  same  meanings  as  in  Art.  83, 


102 


STRENGTH  OF  MATERIALS 


The  point  B  corresponds  very  nearly  to  the  ratio  I  ~-  r  =  90. 
Hence,  for  columns  for  which  the  ratio  Z  -r-  r  is  less  than  90,  the 
first  formula  applies;  and  for  columns  for  which  the  ratio  is 
greater  than  90,  the  second  one  applies.  The  point  C  corre- 
sponds to  the  ratio  Z  -r-  r  =  200,  and  the  second  formula  does  not 
apply  to  a  column  for  which  I  -f-  r  is  greater  than  that  limit. 

P-J-A 

50000 


10OOO 


The  ratio  I  -r-  r  for  steel  columns  of  practice  rarely  Exceeds  150, 
and  is  usually  less  than  100. 

Fig.  53  is  a  combination  of  Figs.  49,  50,  51  and  52,  and 
represents  graphically  a  comparison  of  the  Rankine,  straight-line, 
Euler,  parabola -Euler,  and  broken  straight-line  formulas  for  flat- 
ended  niild-steel  columns,  It  well  illustrates  the  fact  that  our 
knowledge  of  the  strength  of  columns  is  not  so  exact  as  that,  for 
example,  of  the  strength  of  beams. 


100 


200 


3OO 


Fig.  53. 


89.     Design  of  Columns.     All  the  preceding  examples  relat- 
ing to  columns  were  on  either  (1)  computing  the  factor  of  safety 


STRENGTH  OF  MATERIALS  103 

of  a  given  loaded  column,  or  (2)  computing  the  safe  load  for  a 
given  column.  A  more  important  problem  is  to  design  a  column 
to  sustain  a  given  load  under  given  conditions.  A  complete  dis- 
cussion of  this  problem  is  given  in  a  later  paper  on  design.  "We 
show  here  merely  how  to  compute  the  dimensions  of  the  cross  - 
section  of  the  column  after  the  form  of  the  cross-section  has  been 
decided  upon. 

In  only  a  few  cases  can  the  dimensions  be  computed  directly 
(see  example  1  following),  but  usually,  when  a  column  formula  is 
applied  to  a  certain  case,  there  will  be  two  unknown  quantities  in 
it,  A  and  r  or  d.  Such  cases  can  best  be  solved  by  trial  (see 
examples  2  and  3  below). 

Example.  1.  What  is  the  proper  size  of  white  pine  column 
to  sustain  a  load  of  80,000  pounds  with  a  factor  of  safety  of  5, 
when  the  length  of  the  column  is  22  feet  ? 

We  use  the  parabola  formula  (equation  13).  Since  the  safe 
load  is  80,000  pounds  and  the  factor  of  safety  is  5,  the  breaking 
load  P  is 

80,000  X  5  =  400,000  pounds. 

The  unknown  side  of  the  (square)  cross-section  being  denoted  by 
d,  the  area  A  is  d2.  Hence,  substituting  in  the  formula,  since  I 
=  22  feet  =  264  inches,  we  have 

=2,500-  0.6 

Multiplying  both  sides  by  d2  gives 

400,000  =  2,500  d2  -  0.6  X  2642, 

or  2,500  d*  =  400,000  +  0.6  x  264*  =  441,817.6. 

Hence  d2  =  176.73,  or  d  =  13.3  inches. 

2.  What  size  of  cast-iron  column  is  needed  to  sustain  a  load 
of  100,000  pounds  with  a  factor  of  safety  of  10,  the  length  of  the 
column  being  14  feet  ? 

We  shall  suppose  that  it  has  been  decided  to  make  the  cross- 
section  circular,  and  shall  compute  by  Rankine's  formula  modified 
for  cast-iron  columns  (equation  10').  The  breaking  load  for  the 
column  would  be 


104  STRENGTH  OF  MATERIALS 

100,000  X  10  =  1,000,000  pounds. 

The  length  is  14  feet  or  168  inches;  hence  the  formula  oecomes 
1,000,000         80,000 


or,  reducing  by  dividing  both  sides  of  the  equation  by  10,000,  and 
then  clearing  of  fractions,  we  have 

100 

There  are  two  unknown  quantities  in  this  equation,  d  and  A,  and 
we  cannot  solve  directly  for  them.  Probably  the  best  way  to  pro- 
ceed is  to  assume  or  guess  at  a  practical  value  of  d,  then  solve  for 
A,  and  finally  compute  the  thickness  or  inner  diameter.  Thus,  let 
us  try  d  equal  to  7  inches,  first  solving  the  equation  for  A  as  far 
as  possible.  Dividing  both  sides  by  8  we  have 

100  n        1682 

A       .        _  i      I 


and,  combining, 

441 
A  =  12.5  +  ~. 

Now,  substituting  7  for  d,  we  have 

441 
A  =  12.5  +  -JQ-  =  21.5  square  inches. 

The  area  of  a  hollow  circle  whose  outer  and  inner  diameters  are 
d  and  dl  respectively,  is  0.7854  (d2  -  d*).  Hence,  to  find  the  inner 
diameter  of  the  column,  we  substitute  7  for  d  in  the  last  expres- 
sion, equate  it  to  the  value  of  A  just  found,  and  solve  for  dt.  Thus, 

0.7854  (49-^)  =  21.5- 
hence 


and  d*  =  49  -  27.37  =  21.63  or  dl  =  4.65. 

This  value  of  d  makes  the  thickness  equal  to 

J  (7-4.65)  =  1.175  inches, 


STRENGTH  OF  MATERIALS  105 

which  is  safe.     It  might   be  advisable  in  an  actual  case  to  try 
d  equal  to  8  repeating  the  computation.* 

EXAMPLE  FOR  PRACTICE. 

1.  What  size  of  white  oak  column  is  needed  to  sustain  a  load 
of  45,000  pounds  with  a  factor  of  safety  of  6,  the  length  of  the 
column  being  12  feet. 

Ans.     d  =  8-J,  practically  a  10  X  10-inch  section 

STRENGTH  OF  SHAFTS. 

A  shaft  is  a  part  of  a  machine  or  system  of  machines,  and  is 
used  to  transmit  power  by  virtue  of  its  torsional  strength,  or  resist- 
ance  to  twisting.  Shafts  are  almost  always  made  of  metal  and  are 
usually  circular  in  cross-section,  being  sometimes  made  hollow. 

90.  Twisting  Moment.  Let  AF,  Fig.  54,  represent  a  shaft 
with  four  pulleys  on  it.  Suppose  that  D  is  the  driving  pulley 
and  that  B,  C  and  E  are  pulleys  from  which  power  is  taken  off  to 
drive  machines.  The  portions  of  the  shafts  between  the  pulleys 


Pig.  54. 


are  twisted  when  it  is  transmitting  power;  and  by  the  twisting 
moment  at  any  cross-section  of  the  shaft  is  meant  the  algebraic 
sum  of  the  moments  of  all  the  forces  acting  on  the  shaft  on  either 


*NOTE.  The  structural  steel  handbooks  contain  extensive  tables  by 
means  of  which  the  design  of  columns  of  steel  or  cast  iron  is  much  facilitated. 
The  difficulties  encountered  in  the  use  of  formulae  are  well  illustrated  in  this 
example. 


106  STRENGTH  OF  MATERIALS 

side  of  the  section,  the  moments  being  taken  with  respect  to  the 
axis  of  the  shaft.  Thus,  if  the  forces  acting  on  the  shaft  (at  the 
pulleys)  are  P15  P2,  P3,  and  P4  as  shown,  and  if  the  arms  of  the 
forces  or  radii  of  the  pulleys  are  al9  az,  #3,  and  a^  respectively,  then 
the  twisting  moment  at  any  section  in 

BC  is  Pj  «„ 

CD  is  P,  a,  +  P2  «a, 

DE  is  Pj  a,  +  P2  a2  -  P3  a3. 

Like  bending  moments,  twisting  moments  are  usually  ex- 
pressed  in  inch -pounds. 

Example.  Let  at  =  a2  =  a4  =  15  inches,  a3  ==  30  inches, 
P,  =  400  pounds,  P2  =  500  pounds,  P3  =  750  pounds,  and  P4  = 
600  pounds.*  What  is  the  value  of  the  greatest  twisting  moment 
in  the  shaft  ? 

At  any  section  between  the  first  and  second  pulleys,  the 
twisting  moment  is 

400  X  15  =  6,000  inch-pounds; 

at  any  section  between  the  second  and  third  it  is 

400  X  15  +  500  X  15  =  13,500  inch-pounds;  and 
at  any  section  between  the  third  and  fourth  it  is 

400  X  15  +  500  X  15  -  750  X  30  =  -  9,000  inch-pounds. 
Hence  the  greatest  value  is  13,500  inch-pounds. 

91.  Torsional  Stress.  The  stresses  in  a  twisted  shaft  are 
called  "torsional"  stresses.  The  torsional  stress  on  a  cross -section 
of  a  shaft  is  a  shearing  stress,  as  in  the  case  illustrated  by  Fig.  55, 
which  represents  a  flange  coupling  in  a  shaft.  Were  it  not  for 
the  bolts,  one  flange  would  slip  over  the  other  when  either  part 
of  the  shaft  is  turned;  but  the  bolts  prevent  the  slipping.  Obvi- 
ously there  is  a  tendency  to  shear  the  bolts  off  unless  they  are 
screwed  up  very  tight;  that  is,  the  material  of  the  bolts  is  sub- 
jected to  shearing  stress. 

Just  so,  at  any  section  of  the  solid  shaft  there  is  a  tendency 
for  one  part  to  slip  past  the  other,  and  to  prevent  the  slipping  or 

*  Note.  These  numbers  were  so  chosen  that  the  moment  of  P  (driving 
moment)  equals  the  sum  of  the  moments  of  the  other  forces.  This  is  always 
the  case  in  a  shaft  rotating  at  constant  speed;  that  is,  the  power  given  the 
shaft  equals  the  power  taken  off. 


STRENGTH  OF  MATERIALS 


107 


shearing  of  the  shaft,  there  arise  shearing  stresses  at  all  parts  of 
the  cross -section.  The  shearing  stress  on  the  cross-section  of  a 
shaft  is  not  a  uniform  stress,  its  value  per  unit-area  being  zero  at 
the  center  of  the  section,  and  increasing  toward  the  circumference. 
In  circular  sections,  solid  or  hollow,  the  shearing  stress  per  unit- 
area  (unit-stress)  varies  directly  as  the  distance  from  the  center 
of  the  section,  provided  the  elastic  limit  is  not  exceeded.  Thus, 
if  the  shearing  unit-stress  at  the  circumference  of  a  section  is 


Fig.  55. 

1,000  pounds  per  square  inch,  and  the  diameter  of  the  shaft  is 
2  inches,  then,  at  -J  inch  from  the  center,  the  unit-stress  is  500 
pounds  per  square  inch;  and  at  J  inch  from  the  center  it  is  250 
pounds  per  square  inch.  In  Fig.  55  the  arrows  indicate  the 
values  and  the  directions  of  the  shearing  stresses  on  very  small 
portions  of  the  cross-section  of  a  shaft  there  represented. 

92.  Resisting  Moment.      By  "resisting  moment"  at  a  sec- 
tion of  a  shaft  is  meant  the  sum  of  the  moments  of  the  shearing 
stresses  on  the  cross- section  about  the  axis  of  the  shaft. 

Let  Ss  denote  the  value  of  the  shearing  stress  per  unit-area 
(unit-stress)  at  the  outer  points  of  a  section  of  a  shaft;  d  the 
diameter  of  the  section  (outside  diameter  if  the  shaft  is  hollow); 
and  dl  the  inside  diameter.  Then  it  can  be  shown  that  the  re- 
sisting moment  is: 

For  a  solid  section,    0.1963  Ss  d3; 

0.1963  Ss  (tf  -  df) 

lor  a  hollow  section,    * 

d> 

93.  Formula  for  the  Strength  of  a  Shaft.     As  in  the  case 


108  STRENGTH  OF  MATERIALS 

of  beams,  the  resisting  moment  equals  the  twisting  moment  at 
any  section.  If  T  be  used  to  denote  twisting  moment,  then  we 
have  the  formulas  : 

For  solid  circular  shafts,  0.1963  Ss  d?  =  T; 

w    VTI         •      i  -    -u  **     0.1963  Ss  (d*  -  d*)      r 
For  hollow  circular  shafts,  -  J-^  -  L^  = 

d  } 

In  any  portion  of  a  shaft  of  constant  diameter,  the  unit- 
shearing  stress  Ss  is  greatest  where  the  twisting  moment  is  greatest. 
Hence,  to  compute  the  greatest  unit-shearing  stress  in  a  shaft, 
we  first  determine  the  value  of  the  greatest  twisting  moment, 
substitute  its  value  in  the  first  or  second  equation  above,  as  the 
case  may  be,  and  solve  for  Ss.  It  is  customary  to  express  T  in 
inch-pounds  and  the  diameter  in  inches,  Ss  then  being  in  pounds 
per  square  inch. 

Examples.  1.  Compute  the  value  of  the  greatest  shearing 
unit-stress  in  the  portion  of  the  shaft  between  the  first  and  second 
pulleys  represented  in  Fig.  54,  assuming  values  of  the  forces  and 
pulley  radii  as  given  in  the  example  of  article  90.  Suppose  also 
that  the  shaft  is  solid,  its  diameter  being  2  inches. 

The  twisting  moment  T  at  any  section  of  the  portion  between 
the  first  and  second  pulleys  is  6,000  inch-pounds,  as  shown  in  the 
example  referred  to.  Hence,  substituting  in  the  first  of  the  two 
formulas  15  above,  we  have 

0.1963  Ss  X  23  ==  6,000; 

fi  000 

or,  Ss  =  n  1Q'      xx  o  =  3,820  pounds  per  square  inch. 

/\  o 


This  is  the  value  of  the  unit-stress  at  the  outside  portions  of  all 
sections  between  the  first  and  second  pulleys. 

2.  A  hollow  shaft  is  circular  in  cross-section,  and  its  outer 
and  inner  diameters  are  16  and  8  inches  respectively.  If  the 
working  strength  of  the  material  in  shear  is  10,000  pounds  per 
square  inch,  what  twisting  moment  can  the  shaft  safely  sustain  ? 

The  problem  requires  that  we  merely  substitute  the  values  of 
Ss,  <?,  and  d^  in  the  second  of  the  above  formulas  15,  and  solve  for 
T.  Thus, 

rp         0.1963  X  10,000  (16*  -  84)        -  ™  Q0n  .     , 

-  -  JL_          —  —  L  =  7,537,920  inch-pounds. 


STRENGTH  OF  MATERIALS  109 

EXAMPLES  FOR  PRACTICE. 

1.  Compute  the  greatest  value  of  the  shearing  unit-stress  in 
the  shaft  represented  in  Fig.  54,  using  the  values  of  the  forces 
and  pulley  radii  given  in  the  example  of  article  90,  the  diameter 
of  the  shaft  being  2  inches. 

Ans.     8,595  pounds  per  square  inch 

2.  A  solid  shaft  is  circular  in  cross-section  and  is  9.6  inches 
in  diameter.     If  the  working  strength  of  the  material  in  shear  is 
10,000  pounds  per  square  inch,  how  large  a  twisting  moment  can 
the  shaft  safely  sustain?     (The  area  of  the  cross  -section  is  practically 
the  same  as  that  of  the  hollow  shaft  of  example  2  preceding.) 

Ans.  1,736,736  inch-pounds. 

94.  Formula  for  the  Power  Which  a  Shaft  Can  Transmit. 
The  power  that  a  shaft  can  safely  transmit  depends  on  the  shear- 
ing working  strength  of  the  material  of  the  shaft,  on  the  size  of 
the  cross-section,  and  on  the  speed  at  which  the  shaft  rotates. 

Let  H  denote  the  amount  of  horse  -power;  Ss  the  shearing 
working  strength  in  pounds  per  square  inch;  d  the  diameter 
(outside  diameter  if  the  shaft  is  hollow)  in  inches;  dl  the  inside 
diameter  in  inches  if  the  shaft  is  hollow;  and  n  the  number  of 
revolutions  of  the  shaft  per  minute.  Then  the  relation  between 
power  transmitted,  unit-stress,  etc.,  is: 

For  solid  shafts,    H  =  321  QOQ  '*' 

L       (16) 

O  ' 

For  hollow  shafts,  H  = 


.  1.  What  horse-power  can  a  hollow  shaft  16 
inches  and  8  inches  in  diameter  safely  transmit  at  50  revolutions 
per  minute,  if  the  shearing  working  strength  of  the  material  is 
10,000  pounds  per  square  inch? 

We  have  merely  to  substitute  in  the  second  of  the  two  for- 
mulas  16  above,  and  reduce.  Thus, 

II  =          321  QOQ  X  16  --  =  6j0°°  horse-Power  (nearlj)- 

2.  What  size  of  solid  shaft  is  needed  to  transmit  6,000  horse- 
power  at  50  revolutions  per  minute  if  the  shearing  working 
strength  of  the  material  is  10,000  pounds  per  square  inch? 


110  STRENGTH  OF  MATERIALS 

We  have  merely  to  substitute  in  the  first  of  the  two  formulas 
16,  and  solve  for  d.     Thus, 

10,000  X  <P  X  50 
6'0(  321,000       "' 

6,000  X  321,000 
therefore  df          1U)00o  x  50        =  3>852; 


or,  d  =      3852  =  15.68  inches. 

(A  solid  shaft  of  this  diameter  contains  over  25%  more  material  than 
the  hollow  shaft  of  example  1  preceding.  There  is  therefore  considerable 
saving  of  material  in  the  hollow  shaft.) 

3.  A  solid  shaft  4  inches  in  diameter  transmits  200  horse- 
power while  rotating  at  200  revolutions  per  minute.  "What  is  the 
greatest  shearing  unit-stress  in  the  shaft? 

We  have  merely  to  substitute  in  the  first  of  the  equations  16, 
and  solve  for  Ss.  Thus, 

X  43  X   200 


200  = 


321,000 


200  X  321,000 
or,     S  = 43  \/  200 =  ">.    "  pounds  per  square  inch. 

EXAMPLES  FOR  PRACTICE. 

1.  What  horse-power  can  a  solid  shaft  9.6  inches  in  diameter 
safely  transmit  at  50  revolutions  per  minute,  if  its  shearing  work- 
ing  strength  is  10,000  pounds  per  square  inch  ? 

Ans.     1,378  horse-power. 

2.  What  size  of  solid  shaft  is  required  to  transmit  500  horse- 
power at  150  revolutions  per  minute,  the  shearing  working  strength 
of  the  material  being  8,000  pounds  per  square  inch. 

Ans.     5.1  inches. 

3.  A  hollow  shaft  whose  outer  diameter  is  14  and  inner  6.7 
inches  transmits  5,000  horse-power  at  60  revolutions  per  minute. 
What  is  the  value  of  the  greatest  shearing  unit-stress  in  the  shaft? 

Ans.     10,273  pounds  per  square  inch. 

STIFFNESS  OF  RODS,  BEAMS,  AND  SHAFTS. 

The   preceding   discussions    have  related  to  the  strength  of 


STRENGTH  OF  MATERIALS  111 

materials.     We  shall  now  consider  principally  the  elongation  of 
rods,  deflection  of  beams,  and  twist  of  shafts. 

95.  Coefficient  of  Elasticity.  According  to  Hooke's  Law 
(Art.  9,  p.  7),  the  elongations  of  a  rod  subjected  to  an  increasing 
pull  are  proportional  to  the  pull,  provided  that  the  stresses  due  to 
the  pull  do  not  exceed  the  elastic  limit  of  the  material.  Within 
the  elastic  limit,  then,  the  ratio  of  the  pull  and  the  elongation  is 
constant;  hence  the  ratio  of  the  unit-stress  (due  to  the  pull)  to  the 
unit-elongation  is  also  constant.  This  last-named  ratio  is  called 
"  coefficient  of  elasticity."  If  E  denotes  this  coefficient,  S  the 
unit-stress,  and  s  the  unit-deformation,  then 

«-T>  07) 

Coefficients  of  elasticity  are  usually  expressed  in  pounds  per  square  inch. 

The  preceding  remarks,  definition,  and  formula  apply  also  to 
a  case  of  compression,  provided  that  the  material  being  compressed 
does  not  bend,  but  simply  shortens  in  the  direction  of  the  com- 
pressing forces.  The  following  table  gives  the  average  values  of 
the  coefficient  of  elasticity  for  various  materials  of  construction: 

TABLE  Q. 

Coefficients  of  Elasticity. 


Material. 

Average  Coefficient  of  Elasticity. 

Steel  

30,000,000  pounds  per  square  inch. 

Wrought  iron  
Cast  iron  

27,500,000 
15,000,000        "         "         "           " 

Timber  . 

1,800,000        "         "         "           " 

The  coefficients  of  elasticity  for  steel  and  wrought  iron,  for  different 
grades  of  those  materials,  are  remarkably  constant;  but  for  different  grades 
of  cast  iron  the  coefficients  range  from  about  10,000,000  to  30,000,000  pounds 
per  square  inch.  Naturally  the  coefficient  has  not  the  same  value  for  the 
different  kinds  of  wood;  for  the  principal  woods  it  ranges  from  1,600,000 
(for  spruce)  to  2,100,000  (for  white  oak). 

Formula  17  can  be  put  in  a  form  more  convenient  for  use,  as 
follows  : 

Let  P  denote  the  force  producing  the  deformation  ;  A  the 
area  of  the  cross -section  of  the  piece  on  which  P  acts  ;  I  the  length 
of  the  piece  ;  and  D  the  deformation  (elongation  or  shortening). 


11?  STRENGTH  OF  MATERIALS 

Then 

S  =  P  -j-  A  (see  equation  1), 
and  s  =  D  -+-  I  (see  equation  2). 

Hence,  substituting  these  values  in  equation  17,  we  have 


d7') 

The  first  of  these  two  equations  is  used  for  computing  the  value  of 
the  coefficient  of  elasticity  from  measurements  of  a  "  test,"  and 
the  second  for  computing  the  elongation  or  shortening  of  a  given 
rod  or  bar  for  which  the  coefficient  is  known. 

Examples.  1.  It  is  required  to  compute  the  coefficient  of 
elasticity  of  the  material  the  record  of  a  test  of  which  is  given  on 
page  9. 

Since  the  unit-stress  S  and  unit-elongation  s  are  already 
computed  in  that  table,  we  can  use  equation  17  instead  of  the  first 
of  equations  17'.  The  elastic  limit  being  between  40,000  and 
45,000  pounds  per  square  inch,  we  may  use  any  value  of  the 
unit-stress  less  than  that,  and  the  corresponding  unit-elongation. 

Thus,  with  the  first  values  given, 

5,000 


With  the  second, 


This  lack  of  constancy  in  the  value  of  E  as  computed  from  different 
loads  in  a  test  of  a  given  material,  is  in  part  due  to  errors  in  measuring  the 
deformation,  a  measurement  difficult  to  make.  The  value  of  the  coefficient 
adopted  from  such  a  test,  is  the  average  of  all  the  values  of  E  which  can  be 
computed  from  the  record. 

2.     How  much  will  a  pull  of  5,000  pounds  stretch  a  round 
steel  rod  10  feet  long  and  1  inch  in  diameter  ? 

We  use  the  second  of  the  two   formulas   17'.     Since  A  = 
0.7854  X  I2  ==  0.7854  square  inches,  I  ?*  120  inches,  and  E  - 
30,000,000  pounds  per  square  inch,  the  stretch  is: 


STRENGTH  OF  MATERIALS  113 

EXAMPLES  FOR  PRACTICE. 

1.  "What  is  the  coefficient  of  elasticity  of  a  material  if  a  pull 
of  20,000  pounds  will  stretch  a  rod  1  inch  in  diameter  and  4  feet 

long  0.045  inch  ? 

Ans.  27,000,000  pounds  per  square  inch. 

2.  How  much  will  a  pull  of  15,000  pounds  elongate  a  round 
cast-iron  rod  10  feet  long  and  1  inch  in  diameter  ? 

Ans.  0.152  inch. 

96.  Temperature  Stresses.  In  the  case  of  most  materials, 
when  a  bar  or  rod  is  heated,  it  lengthens;  and  \vhen  cooled,  it 
shortens  if  it  is  free  to  do  so.  The  coefficient  of  linear  expansion 
of  a  material  is  the  ratio  which  the  elongation  caused  in  a  rod  or 
bar  of  the  material  by  a  change  of  one  degree  in  temperature  bears 
to  the  length  of  the  rod  or  bar.  Its  values  for  Fahrenheit  degrees 
are  about  as  follows: 

For  Steel,  0.0000065. 

For  Wrought  iron,    .0000067. 
For  Cast  iron,  .0000062. 

Let  K  be  used  to  denote  this  coefficient;  t  a  change  of  tem- 
perature, in  degrees  Fahrenheit;  I  the  length  of  a  rod  or  bar; 
and  D  the  change  in  length  due  to  the  change  of  temperature. 
Then 

D  =  =  K  tl.  (18) 

D  and  Z  are  expressed  in  the  same  unit. 

If  a  rod  or  bar  is  confined  or  restrained  so  that  it  cannot 
change  its  length  when  it  is  heated  or  cooled,  then  any  change  in 
its  temperature  produces  a  stress  in  the  rod;  such  are  called  tem- 
perature stresses. 

Examples.  1.  A  steel  rod  connects  two  solid  walls  and  is 
screwed  up  so  that  the  unit-stress  in  it  is  10,000  pounds  per 
square  inch.  Its  temperature  falls  10  degrees,  and  it  is  observed 
that  the  walls  have  not  been  drawn  together.  What  is  the  temper- 
ature stress  produced  by  the  change  of  temperature,  and  what  is 
the  actual  unit-stress  in  the  rod  at  the  new  temperature  ? 

Let  I  denote  the  length  of  the  rod.  Then  the  change  in 
length  which  would  occur  if  the  rod  were  free,  is  given  by  formula 
18,  above,  thus: 

D  =  0.0000065  X  10  X  I  =  0.000065  I. 


114  STRENGTH  OF  MATERIALS 

Now,  since  the  rod  could  not  shorten,  it  has  a  greater  than  normal 
length  at  the  new  temperature;  that  is,  the  fall  in  temperature  has 
produced  an  effect  equivalent  to  an  elongation  in  the  rod  amount- 
ing to  D,  and  hence  a  tensile  stress.  This  tensile  stress  can  be 
computed  from  the  elongation  D  by  means  of  formula  17.  Thus, 

S  =  Es; 
and  since  s,  the  unit-elongation,  equals 

D  =    .0000065  I  =  JMOO(ft 

i  i 

S  =  30,000,000  X    .0000065  =  195.0  pounds  per  square  inch. 
This  is  the  value  of  the  temperature  stress;  and  the  new  unit- 
stress  equals 

10,000  +  195.0  =  10,195  pounds  per  square  inch. 

Notice  that  the  unit  temperature  stresses  are  independent  of  the  length 
of  the  rod  and  the  area  of  its  cross-section. 

2.  Suppose  that  the  change  of  temperature  in  the  preceding 
example  is  a  rise  instead  of  a  fall.  What  are  the  values  of  the 
temperature  stress  due  to  the  change,  and  of  the  new  unit- stress  in 
the  rod  ? 

The  temperature  stress  is  the  same  as  in  example  1,  that  is, 
1,950  pounds  per  square  inch ;  but  the  rise  in  temperature 
releases,  as  it  were,  the  stress  in  the  rod  due  to  its  being  screwed 
up,  and  the  final  unit  stress  is 

10,000  -  1,950  =  8,050  pounds  per  square  inch. 

EXAHPLE  FOR  PRACTICE, 

1.     The  ends  of  a  wrought-iron  rod  1  inch  in  diameter  are 

fastened  to  two  heavy  bodies  which  are  to  be  drawn  together,  the 
j  & 

temperature  of  the  rod  being  200  degrees  when  fastened  to  the  ob- 
jects. A  fall  of  120  degrees  is  observed  not  to  move  them. 
What  is  the  temperature  stress,  and  what  is  the  pull  exerted  by 
the  rod  on  each  object  ? 

(  Temperature  stress,  22,000  pounds  per  square  inch. 
AnS'  \  Pull,  17,280  pounds. 

97.  Deflection  of  Beams.  Sometimes  it  is  desirable  to  know 
how  much  a  given  beam  will  deflect  under  a  given  load,  or  to  design 


STRENGTH  OF  MATERIALS  115 

a  beam  which  will  not  deflect  more  than  a  certain  amount  under  a 
given  load.  In  Table  B,  page  53,  Part  I,  are  given  formulas  for 
deflection  in  certain  cases  of  beams  and  different  kinds  of  loading. 

In  those  formulas,  d  denotes  deflection;  I  the  moment  of  inertia  of  the 
cross-section  of  the  beam  with  respect  to  the  neutral  axis,  as  in  equation  6 ; 
and  E  the  coefficient  of  elasticity  of  the  material  of  the  beam  (for  values,  see 
Art.  95). 

In  each  case,  the  load  should  be  expressed  in  pounds,  the  length  in 
inches,  and  the  moment  of  inertia  in  biquadratic  inches;  then  the  deflection 
will  be  in  inches. 

According  to  the  formulas  for  d,  the  deflection  of  a  beam 
varies  inversely  as  the  coefficient  of  its  material  (E)  and  the  mo- 
ment of  inertia  of  its  cross-section  (I)  ;  also,  in  the  first  four  and 
last  two  cases  of  the  table,  the  deflection  varies  directly  as  the  cube 
of  the  length  (Z3). 

Example.  What  deflection  is  caused  by  a  uniform  load  of 
6,400  pounds  (including  weight  of  the  beam)  in  a  wooden  beam 
on  end  supports,  which  is  12  feet  long  and  6  X  12  inches  in 
cross -section  ?  (This  is  the  safe  load  for  the  beam  ;  see  example 
1,  Art.  65.) 

The  formula  for  this  case  (see  Table  B,  page  53)  is 

5  W 
=  384  El ' 

Here  W  =  6,400  pounds  ;  I  =  144  inches  ;  E  =  1,800,000 
pounds  per  square  inch  ;  and 

I  =  -^  la?  =  jg-  6  X  123=  864  inches4. 

Hence  the  deflection  is 

5  X  6,400  X  144* 
=  384  X  1,800,000  X  864  = 

EXAMPLES  FOR  PRACTICE. 

1.  Compute  the  deflection  of  a  timber  built-in   cantilever 
8X8  inches  which  projects  8  feet  from  the  wall  and  bears  an 
end  load  of  900  pounds.     (This  is  the  safe  load  for  the  cantilever, 
see  example  1,  Art.  65.) 

Ans.     0.43  inch. 

2.  Compute  the  deflection  caused  by  a  uniform  load  of  40,000 


116 


STRENGTH  OF  MATERIALS 


pounds  on  a  42-pound  15-inch  steel  I-beam  which  is  16  feet  long 
and  rests  on  end  supports. 

Ans.     0.28  inch. 

98.     Twist  of  Shafts.     Let  Fig.  57  represent  a  portion  of  a 
shaft,    and  suppose  that  the  part  represented  lies  wholly  between 


Fig.  57. 

two  adjacent  pulleys  on  a  shaft  to  which  twisting  forces  are  applied 
(see  Fig.  54).  Imagine  two  radii  ma  and  nb  in  the  ends  of  the 
portion,  they  being  parallel  as  shown  when  the  shaft  is  not  twisted. 
After  the  shaft  is  twisted  they  will  not  be  parallel,  ma  having 
moved  to  ma',  and  nb  to  nb1 '.  The  angle  between  the  two  lines  in 
their  twisted  positions  (ma'  and  nb')  is  called  the  angle  of  twist, 
or  angle  of  torsion,  for  the  length  1.  If  a  a"  is  parallel  to  ab,  then 
the  angle  a"nb'  equals  the  angle  of  torsion. 

If  the  stresses  in  the  portion  of  the  shaft  considered  do  not 
exceed  the  elastic  limit,  and  if  the  twisting  moment  is  the  same 
for  all  sections  of  the  portion,  then  the  angle  of  torsion  a  (in 
degrees)  can  be  computed  from  the  following: 

For  solid  circular  shafts, 


a  = 


584  TZ      36,800,000  HI 


For  hollow  circular  shafts, 


(19) 


584  Tld        36,800,000  HZ 


Here  T,  Z,  d,  dl9  H,  and  n  have  the  same  meanings  as  in  Arts.  93 
and  94,  and  should  be  expressed  in  the  units  there  used.  The 
letter  E1  stands  for  a  quantity  called  coefficient  of  elasticity  for 
shear;  it  is  analogous  to  the  coefficient  of  elasticity  for  tension  and 
compression  (E),  Art.  95.  The  values  of  E1  for  a  few  materials 
average  about  as  follows  (roughly  E1  =  |  E) : 


STRENGTH  OF  MATERIALS  117 

For  Steel,  11,000,000  pounds  per  square  inch. 

For  Wrought  iron,  10,000,000        "          "         "          " 
For  Cast  iron,  6,000,000        "          "         «  " 

Example.  What  is  the  value  of  the  angle  of  torsion  of  a 
steel  shaft  60  feet  long  when  transmitting  6,000  horse-power  at 
50  revolutions  per  minute,  if  the  shaft  is  hollow  and  its  outer  and 
inner  diameters  are  16  and  8  inches  respectively  ? 

Here  I  —  720  inches;  hence,  substituting  in  the  appropriate 
formula  (19),  we  find  that 


36,800,000  X  6,000  x 
=  11,000,000  X  (16-  -  80  50  = 

EXAMPLE  FOR  PRACTICE. 

Suppose  that  the  first  two  pulleys  in  Fig.  54  are  12  feet 
apart;  that  the  diameter  of  the  shaft  is  2  inches;  and  that  P,  =  400 
pounds,  and  al  =  15  inches.  If  the  shaft  is  of  wrought  iron, 
what  is  the  value  of  the  angle  of  torsion  for  the  portion  between 
the  first  two  pulleys  ? 

Ans.     3.15  degrees. 

99.  Non-elastic  Deformation.  The  preceding  formulas  for 
elongation,  deflection,  and  twist  hold  only  so  long  as  the  greatest 
unit-stress  does  not  exceed  the  elastic  limit.  There  is  no  theory, 
and  no  formula,  for  non  -elastic  deformations,  those  corresponding 
to  stresses  which  exceed  the  elastic  limit.  It  is  well  known,  how- 
ever, that  non  -elastic  deformations  are  not  proportional  to  the 
forces  producing  them,  but  increase  much  faster  than  the  loads. 
The  value  of  the  ultimate  elongation  of  a  rod  or  bar  (that  is,  the 
amount  of  elongation  at  rupture),  is  quite  well  known  for  many 
materials.  This  elongation,  for  eight-inch  specimens  of  various 
materials  (see  Art.  16),  is  : 

For  Cast  iron,  about  1  per  cent. 
For  Wrought  iron  (plates),  12  -  15  per  cent. 
For         "  "        (bars),  20-25    "      "   . 

For  Structural  steel,  22-26    "      "   . 

Specimens  of  ductile  materials  (such  as  wrought  iron  and 
structural  steel),  when  pulled  to  destruction,  neck  down,  that  is, 
diminish  very  considerably  in  cross-section  at  some  place  along 
the  length  of  the  specimen.  The  decrease  in  cross-sectional  area 


118 


STRENGTH  OF  MATERIALS 


is  known  as  reduction  of  area,  and  its  value  for  wrought  iron  and 
steel  may  be  as  much  as  50  per  cent. 

RIVETED  JOINTS. 

100.  Kinds  of  Joints.  A  lap  joint  is  one  in  which  the 
plates  or  bars  joined  overlap  each  other,  as  in  Fig.  58,  a.  A  butt 
joint  is  one  in  which  the  plates  or  bars  that  are  joined  butt  against 
each  other,  as  in  Fig.  58,  b.  The  thin  side  plates  on  butt  joints 


Fig.  58. 

are  called  cover=plates ;  the  thickness  of  each  is  always  made  not 
less  than  one-half  the  thickness  of  the  main  plates,  that  is,  the 
plates  or  bars  that  are  joined.  Sometimes  butt  joints  are  made 
with  only  one  cover-plate;  in  such  a  case  the  thickness  of  the 
cover-plate  is  made  not  less  than  that  of  the  main  plate. 

"When  wide  bars  or  plates  are  riveted  together,  the  rivets  are 
placed  in  rows,  always  parallel  to  the  "  seam  "  and  sometimes  also 
perpendicular  to  the  seam;  but  when  we  speak  of  a  row  of  rivets, 
we  mean  a  row  parallel  to  the  seam.  A  lap  joint  with  a  single 
row  of  rivets  is  said  to  be  single=riveted ;  and  one  with  two  rows 
of  rivets  is  said  to  be  double-riveted.  A  butt  joint  with  two  rowa 
of  rivets  (one  on  each  side  of  the  joint)  is  called  "  single-riveted," 
and  one  with  four  rows  (two  on  each  side)  is  said  to  be  "double- 
riveted." 

The  distance  between  the  centers  of  consecutive  holes  in  a 
row  of  rivets  is  called  pitch. 

101.  Shearing  Strength,  or  Shearing  Value,  of  a  Rivet. 
When  a  lap  joint  ia  subjected  to  tension  (that  is,  when  P,  Fig.  58, 
#,  is  a  pull),  and  when  the  joint  is  subjected  to  compression  (when 
P  is  a  push),  there  is  a  tendency  to  cut  or  shear  each  rivet  along 
the  surface  between  the  two  plates.  In  butt  joints  with  two  cover- 


STRENGTH  OF  MATERIALS  119 

plates,  there  is  a  tendency  to  cut  or  shear  each  rivet  on  two  sur- 
faces (see  Fig.  58,  b).  Therefore  the  rivets  in  the  lap  joint  are 
said  to  be  in  single  shear  ;  and  those  in  the  butt  joint  (two  covers) 
are  said  to  be  in  double  shear. 

The  "  shearing  value  "  of  a  rivet  means  the  resistance  which 
it  can  safely  offer  to  forces  tending  to  shear  it  on  its  cross -section. 
This  value  depends  on  the  area  of  the  cross-section  and  on  the  work- 
ing strength  of  the  material.  Let  d  denote  the  diameter  of  the 
cross-section,  and  Ss  the  shearing  working  strength.  Then,  since 
the  area  of  the  cross -section  equals  0.7854  d2,  the  shearing  strength 
of  one  rivet  is  : 

For  single  shear,  0.7854  d2  S8  . 

For  double  shear,  1.5708  <&  S,  . 

102.  Bearing  Strength,  or  Bearing  Value,  of  a  Plate.     When 
a  joint  is  subjected  to  tension  or  compression,  each  rivet  presses 
against  a  part  of  the  sides  of  the  holes  through  which  it  passes. 
By  "  bearing  value  "  of  a  plate  (in  this  connection)  is  meant  the 
pressure,  exerted  by  a  rivet  against  the  side  of  a  hole  in  the  plate, 
which    the   plate    can   safely  stand.     This  value  depends  on  the 
thickness  of  the  plate,  on  the  diameter  of  the  rivet,  and  on  the 
compressive   working    strength    of   the    plate.     Exactly    how    it 
depends  on  these  three  qualities  is  not  known;  but  the  bearing 
value  is  always  computed  from  the  expression  t  d  Sc,  wherein  t 
denotes  the  thickness  of  the  plate;  6?,  the  diameter  of  the  rivet  or 
hole;  and  Sc,  the  working  strength  of  the  plate. 

103.  Frictional  Strength  of  a  Joint.     When  a  joint  is  sub- 
jected  to  tension  or  compression,  there  is  a  tendency  to  slippage 
between  the  faces  of  the  plates  of  the  joint.     This  tendency  is 
overcome  wholly  or  in  part  by  frictional  resistance  between  the 
plates.     The  frictional    resistance  in  a    well-made  joint  may  be 
very  large,  for  rivets  are  put  into  a  joint  hot,  and  are  headed  or 
capped  before  being  cooled.    In  cooling  they  contract,  drawing  the 
plates  of  the  joint  tightly  against  each  other,  and  producing  a 
great  pressure  between  them,  which  gives  the  joint  a  correspond- 
ingly large  frictional  strength.     It  is  the  opinion  of  some  that 
all  well-made    joints    perform    their  service  by   means    of  their 
frictional  strength;  that  is  to  say,  the  rivets  act  only  by  pressing 
the    plates    together   and    are    not   under    shearing    stress,    nor 


120  STRENGTH  OF  MATERIALS 

are  the  plates  under  compression  at  the  sides  of  their  holes.  The 
"  frictional  strength  "  of  a  joint,  however,  is  usually  regarded  as 
uncertain,  and  generally  no  allowance  is  made  for  friction  in  com- 
putations  on  the  strength  of  riveted  joints. 

104.  Tensile  and  Compressive  Strength  of  Riveted  Plates. 
The  holes  punched  or  drilled  in    a  plate  or  bar  weaken  its  tensile 
strength,  and  to  compute  that  strength  it  is  necessary  to  allow  for 
the  holes.     By  net  section,  in  this  connection,  is  meant  the  small- 
est  cross-section  of  the  plate  or  bar  ;  this  is  always  a  section  along 
a  line  of  rivet  holes. 

If,  as  in  the  foregoing  article,  t  denotes  the  thickness  of  the 
plates  joined  ;  d,  the  diameter  of  the  holes;  nl9  the  number  of  riv- 
ets in  a  row  ;  and  w,  the  width  of  the  plate  or  bar  ;  then  the  net 
section  =  (w  -  ntd)  t. 

Let  St  denote  the  tensile  working  strength  of  the  plate  ;  then 
the  strength  of  the  unriveted  plate  is  wtSi9  and  the  reduced  tensile 
strength  is  (w  -  n^)  t  St. 

The  compressive  strength  of  a  plate  is  also  lessened  by  the 
presence  of  holes  ;  but  when  they  are  again  filled  up,  as  in  a  joint, 
the  metal  is  replaced,  as  it  were,  and  the  compressive  strength  of 
the  plate  is  restored.  No  allowance  is  therefore  made  for  holes  in 
figuring  the  compressive  strength  of  a  plate. 

105.  Computation  of  the  Strength  of  a  Joint.     The  strength 
of  a  joint  is  determined  by  either  (1)  the  shearing  value  of  the 
rivets  ;    (2)   the  bearing  value  of  the  plate  ;  or   (3)  the  tensile 
strength  of  the  riveted  plate  if  the  joint  is  in  tension.     Let  P8  de- 
note  the  strength  of  the  joint  as   computed  from  the  shearing 
values  of  the  rivets  ;  Pc,  that  computed  from  the  bearing  value  of 
the  plates  ;    and  Pt,    the  tensile  strength  of  the  riveted  plates. 
Then,  as  before  explained, 

Pt=  (W  -  n,d)  *St;  J 

P8=  n2  0.7854  ^2S8;  and    V  (20) 


nz  denoting  the  total  number  of  rivets  in  the  joint  ;  and  na  denot- 
ing the  total  number  of  rivets  in  a  lap  joint,  and  one-half  the 
number  of  rivets  in  a  butt  joint. 

Examples.     1.     Two  half-inch  plates  7^  inches  wide  are  con- 


STRENGTH  OF  MATERIALS  121 


nected  by  a  single  lap  joint  double-riveted,  six  rivets  in  two  rows. 
If  the  diameter  of  the  rivets  is  |  inch,  and  the  working  strengths 
are  as  follows  :  St=  12,000,  S§==  7,500,  and  Sc=  15,000  pounds 
per  square  inch,  what  is  the  safe  tension  which  the  joint  can 
transmit  ? 

Here  nl==  3,  n=  6,  and  n3=  6  ;  hence 

Pt=  (7-i-  -  3  X  ~)  X  -i-  X  12,000  ==  31,500  pounds; 

Ps=  6  X   3.7854  X  (-|-)2  X  7,500  =  19,880  pounds  ; 

PA*=  6  X  -5-  X--J-  X  15,000  =  33,750  pounds. 

Since  P8  is  the  least  of  these  three  values,  the  strength  of  the 
joint  depends  on  the  shearing  value  of  its  rivets,  and  it  equals 
19,880  pounds. 

2.  Suppose  that  the  plates  described  in  the  preceding  example 
are  joined  by  means  of  a  butt  joint  (two  cover-plates),  and  12 
rivets  are  used,  being  spaced  as  before.     What  is  the  safe  tension 
which  the  joint  can  bear  ? 

Here  nl  ==  3,  n2  =  12,  and  n3  =  6;  hence,  as  in  the  preced- 
ing example, 

Pt  =  31,500;  and  Pc  =  33,750  pounds;  but 

Ps  =  12  X  0.7854  X  (-|-)2  X   7,500  =  S9,760  pounda 

The  strength  equals  31,500  pounds,  and  the  joint  is  stronger  than 
the  first. 

3.  Suppose  that  in  the  preceding  example  the  rivets  are 
arranged  in  rows   of  two.     What  is  the  tensile  strength  of  the 
joint  ? 

Here  nl  =  2.  n2  —  12,  and  n3  ==  6;  hence,  as  in  the  preced- 
ing example, 

Ps  ==  39,760;  and  Pc  =  33,750  pounds;  but 

Pt  =  (7  -|--2  X  -|-)  ^-  X  12,000  =  36,000  pounds. 

The  strength  equals  33,750  pounds,  and  this  joint  is  stronger  than 
either  of  the  first  two. 


122  STRENGTH  OF  MATERIALS 

EXAMPLES  FOR  PRACTICE. 

Note.     Use  working  strengths  as  in  example  1,  above. 
St  =  12,000,  S8  =  7,500,  and  Sc  =  15,000  pounds  per  square  inch. 

1.  Two  half-inch  plates  5  inches  wide  are  connected  by  a 
lap  joint,  with  two  |-inch  rivets  in   a  row.     "What  is  the  safe 
strength  of  the  joint  ? 

Ans.     6,625  pounds. 

2.  Solve  the  preceding  example  supposing  that  four  |-inch 
rivets  are  used,  in  two  rows. 

Ans.    13,250  pounds. 

3.  Solve  example  1  supposing  that  three  1-inch  rivets  are 
used,  placed  in  a  row  lengthwise  of  the  joint. 

Ans.    17,670  pounds. 

4.  Two  half  -inch  plates  5  inches  wide  are  connected  by  a 
butt  joint  (two  cover-plates),  and   four  |-inch  rivets  are  used,  in 
two  rows.     What  is  the  strength  of  the  joint  ? 

Ans.     11,250  pounds. 

106.  Efficiency  of  a  Joint.  The  ratio  of  the  strength  of  a 
joint  to  that  of  the  solid  plate  is  called,  the  "  efficiency  of  the 
joint."  If  ultimate  strengths  are  used  in  computing  the  ratio, 
then  the  efficiency  is  called  ultimate  efficiency;  and  if  working 
strengths  are  used,  then  it  is  called  working  efficiency.  In  the 
following,  we  refer  to  the  latter.  An  efficiency  is  sometimes  ex- 
pressed as  a  per  cent.  To  express  it  thus,  multiply  the  ratio 
strength  of  joint  --f-  strength  of  solid  plate,  by  100. 

Example.  It  is  required  to  compute  the  efficiencies  of  the 
joints  described  in  the  examples  worked  out  in  the  preceding  article. 

In  each  case  the  plate  is  -|  inch  thick  and  7J  inches  wide; 
hence  the  tensile  working  strength  of  the  solid  plate  is 

7-L  x  _L  x  12,000  =  45,000  pounds. 
Therefore  the  efficiencies  of  the  joints  are  : 


(2)  =  0.70,  or  70  per  cent; 


INDEX 


Page 

Angle  of  torsion 116 

Angle  of  twist 116 

Applications  of  first  beam  formula 59 

Beam  formula 

first 59 

second 73 

Beams 19 

deflection  of 114 

stiffness  of 110 

Bearing  strength  of  a  plate. . 119 

Bending  moment 23,  32 

maximum 40 

Brick,  ultimate  strength  of 15 

Broken  straight-line  formula 101 

Built-up  sections,  moment  of  inertia  of 49 

Butt  joint 118 

Cast  iron 

in  tension v 13 

in  compression 15 

Center  of  gravity  of  an  area , 41 

Center  of  gravity  of  built-up  sections 44 

Centrally  loaded  column 88 

Coefficient  of  elasticity Ill 

for  shear 116 

Coefficient  of  linear  expansion 113 

Column  formulas 

graphical  representation  of 93 

Rankine's 88 

Column  loads,  kinds  of 88 

Columns 

classes  of 86 

cross-sections  of ; 86 

design  of 102 

end  conditions  of 85 

strength  of 85 

Combination  column  formulas 94 

Combined  flexural  and  direct  stress 83 

Compression,  materials  in 13 

Compressive  strength  of  riveted  plates 120 

Continuous  beam .  .  19 


124  INDEX 

Page 

Cover-plates 118 

Deflection  of  beams 114 

Deformation 4 

Design  of 

columns 102 

timber  beams 76 

Direct  stress 79 

Double  shear 119 

Eccentrically  loaded ; 88 

Efficiency  of  a  joint 122 

Elastic  limit 5 

Elasticity , 4 

End  conditions  of  column 85 

Euler  formulas 94 

External  shear 23 

Factor  of  safety 8 

Fibre  stresses 55,  57 

First  beam  formula • 59 

Flexural  stress 79 

Flexure  and  compression 81 

Flexure  and  tension 79 

Formula  for  power  which  shaft  can  transmit 109 

Formula  for  strength  of  shaft 107 

Frictional  strength  of  joint 119 

Graphical  representation  of  column  formulas 93 

Hooke's  law 5 

Horizontal  shear 75 

Inclined  forces 54 

Joint 

efficiency  of 122 

frictional  strength  of 119 

Kinds  of  loads  and  beams 52,  79 

Kinds  of  stress 2,  55 

Lap  joint 118 

Laws  of  strength  of  beams 71 

Linear  expansion,  coefficient  of 113 

Longitudinal  forces 54 

Main  plates 118 

Materials 

in  compression 13 

in  shear 15 

in  tension 11 

Maximum  bending  moment 40 

Maximum  shear 31 

Metals  in  shear 16 

Modulus  of  rupture 72 

Moment  diagrams 36 

Moment  of  a  force .  .                                                              16 


INDEX  125 

Page 

Moment  of  inertia. 46 

of  built-up  sections 49 

of  a  rectangle 48 

unit  of 47 

Moments,  principle  of 17 

Neutral  axis 54 

Neutral  line 54 

Neutral  surface 54 

Non-elastic  deformation 117 

Notation 24,  33 

Parabola-Euler  formulas 98 

Posts 85 

Principle  of  moments 17 

applied  to  areas 42 

Radius  of  gyration 86 

Rankine's  column  formula 88 

Reactions  of  supports 16 

Rectangle,  moment  of  inertia  of 48 

Reduction  formula 48 

Resisting  moment 57, 107 

value  of 58 

Resisting  shear 73 

Restrained  beam 19 

Rivet,  shearing  strength  of 118 

Riveted  joints 118 

Rods,  stiffness  of 110 

Rule  of  signs . .  .23,  32 

Safe  load  of  a  beam 71 

Safe  strength  of  a  beam 71 

Second  beam  formula 73 

Section  modulus 59 

Shafts 

stiffness  of 110 

strength  of 105 

twist  of 116 

twisting  moment  of 105 

Shear  diagrams 27 

Shear  stress 2 

Shearing  strength  of  rivet 118 

Shears,  unit  for , 24 

Signs,  rule  of 23,  32 

Simple  beam 19 

Simple  stress 1 

Single  shear 119 

Steel  in  compression 14 

Steel  in  tension 12 

Stiffness  of  rods,  beams,  and  shafts 110 

Stone,  ultimate  strength  of 15 


126  INDEX 

Page 

Straight-line  formulas 94 

Strength  of  beams 52 

laws  of 71 

Strength  of  columns 85 

Strength  of  joint,  computation  of 120 

Strength  of  shafts 105 

Stress,  kinds  of 2 

Stress-deformation  diagram 6 

Struts 85 

Tables 

bending  moment 53 

coefficients  of  elasticity Ill 

deflection 53 

factors  of  safety 10 

maximum  shear,  values  of 53 

mild  steel  columns,  data  for 95,  99 

moduli  of  rupture 72 

moment  diagrams 53 

moments  of  inertia 52 

properties  of  standard  I-beams 70 

radii  of  gyration 52 

section  moduli 52 

shear  diagrams 53 

Temperature  stresses 113 

Tensile  strength  of  riveted  plates 120 

Tension,  materials  in 11 

Timber 

in  compression 14 

in  shear 15 

in  tension 11 

Timber  beams,  design  of 76 

Torsional  stress 106 

Transverse  forces 52 

Twist  of  shafts 116 

Twisting  moment  of  shaft 105 

Ultimate  strength 6 

Unit-deformation 4 

Unit  of  moment  of  inertia 47 

Unit-stress 3 

Units 33 

Units  for  shears 24 

Value  of  resisting  moment 58 

Wrought-iron 

in  compression 14 

in  tension 12 

Working  strength 

Working  stress 8 


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